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The ninth course in this series goes back to physics specifically looking at electricity and magnetism, where the first physics course looked at classical mechanics. A lot of the course feels like the full course follow up to Intro to EECS’s section on circuits and it probably makes the most use of calculus a course has so far. Following the force diagrams in the first physics course, electricity looks at charges creating fields which require using integration to calculate the total force from force propagated through each section. Additionally, it looks at different methods of storing and converting energy between types which was hinted at in Classical Mechanics. Admittedly, of all of the courses so far, this is probably the one where I still have the most to learn from even the content covered and plan on returning to these notes and other resources. The course on MIT Open CourseWare consists of a good amount of reading and notes from each class, but unfortunately doesn’t contain videos and most of the animations linked to don’t work. Most of the notes are based on the reading’s provided as most of the individual lecture notes were covered in better detail in the readings. Sometimes the lecture notes provided real work examples or explained something better however, and they do a pretty good job of condensing the reading into the most important concepts and functions. Still, the actual class consisted of experiments which could probably add more of a connection to the functions and concepts that it’s hard to get from the reading alone. Still, it’s always neat to learn about features of the world that I often take for granted and the topics in this course have a lot to do with powering machines used in daily life.

MIT Courses #9: Physics 2: Electricity and Magnetism – 8.02 by Physics Department Faculty, Lecturers and Technical Staff

Chapter 1 – Fields

In electromagnetic field theory, electromagnetic fields are the mediators of the interaction between material objects. James Clerk Maxwell came up with this theory which differed from the previous “action at a distance” view, where interaction between two objects doesn’t require any additional mechanisms and the objects exert force on each other transmitted through empty space. This didn’t take into consideration an agent of transmission. With field theory, objects not in direct contact must exert a force on each other through an intervening medium or mechanism in the space between them. Force is transmitted from the first object to the mechanism and from element to element in the space to the second object. The force is transmitted at a finite speed by stresses in the space via the presence of the objects. Field theory replaces “action at a distance” with “action by continuous contact” where contact is provided by a stress or “field” induced in the space between the objects by their presence. Field theory is the foundation of modern approaches to understanding the world around us and the first field theory was classical electromagnetism.

A scalar field is a function that gives a single value of a variable for every point in space. An example shown is temperatures at night across Mars at different coordinates taken by the Mars Global Surveyor. In the two dimensional map, this doesn’t take altitudes into consideration. A scalar field, in general, will provide values for every point in space rather than simply a two-dimensional surface. Three-dimensional scalar fields can be represented by a mathematical function, such as T(r,θ,Δ) where the value is a temperature given coordinates (r,θ,Δ) for a planet. This function is a scalar field and gives a number instead of a vector (which has magnitude and direction). A scalar field could also be used to describe other physical attributes such as atmospheric pressure, but not every quantity, such as wind velocity which requires direction as well.

A scalar field defined by an equation is shown to be represented in a few possible ways. One way is to fix one of the independent variables and show a contour map for the others where curves represent lines of constant values of the functions and a series of these for fixed values of the variable will give a feel for the function’s properties. This allows the function to be represented by a series of two dimensional graphs. Another way is to use color coding for two dimensions with a fixed value for the third. Using colors and shading to show how the values change. The third method shown is to use a relief map, by fixing one dimension and plotting the value of the function at heights off x and y coordinates.

Vectors have both a magnitude and a direction in space and are used to describe quantities such as velocity, momentum, acceleration and force of objects. To describe a system consisting of a large number of objects such as moving water, snow, or rain, a vector needs to be assigned to each object. For falling snowflakes, as snow falls, each snowflake moves in a direction and at any point in time a velocity vector can characterize its movement. Snowflakes are an example of collections of discrete bodies, but for continuous bodies such as fluids, velocity vectors will need to be assigned to every point for every instant in time. This collection is a velocity vector field and it contains information about direction and magnitude for every point in space. Air flow is an example of one such system. A vector field F(x,y,z) can be written as F_xi-hat + F_y(x,y,z)j-hat + F_x(x,y,z)k-hat. Fluid flow are the easiest vector fields to visualize and an example shows a fixed number of particles flowing downward from the center of a cone (source) due to gravity. Particles are created at origin in this example and flow away in a “diverging flow”. A converging flow or “sink” of particles shows particles moving toward each other. In vector calculus, a velocity V is represented v_xi-hat+v_yj-hat+v_zk-hat. A point (x,y,z) is a source if the divergence of V(x,y,z) is greater than zero. Divergence is represented as an upside down Δ symbol called a nabla, ∇ and the partial differential symbol is ∂. So, ∇*V(x,y,z) = ∂v_x/∂_x * + ∂v_y/∂_y + ∂v_z/∂_z > 0 where &nabla= ∂/∂_x * i-hat + ∂/∂_y * k-hat + ∂/∂_z * k-hat is the del operator. (x,y,z) is a sink if the divergence is less than zero. When it’s zero, then the point is neither a source or sink and a fluid with zero divergence is incompressible. This kind of flow field can exhibit circulation. In an animation, particles aren’t created or destroyed throughout and move in a circular motion.

Flow fields can have more than one systems of circulation around different points in space. Non-zero divergence means that particles are created or destroyed and a flow with curl means particles move in circles. Vector fields can be written as the sum of a curl-free part and divergence-free part. Electrostatic fields are curl free and magnetic fields are divergence free. Electric fields with both a divergence and a curl will be found when looking at time-varying situations, though magnetic fields always remain divergence free. Comparing fluid fields and electromagnetic fields, vector fields representing fluid flow have immediate physical interpretation and the vector at each point represents a direction of motion. The motions can be shown using animation by combining snapshots in different points in time. A general vector field used in electric and magnetic fields don’t have the immediate physical interpretation and there isn’t a “flow” of a fluid in electric or magnetic fields. Many terms that describe fluid flow can describe electromagnetic fields, such as “flux” which would be the amount of liquid flowing across a surface in a unit of time and used to describe electromagnetic fields even without the same physical reality. The same is true for circulation even if no fluid circulates in a magnetic field. These terms are used to help describe electromagnetic fields to understand their structure even if the analogy is limited.

The gravitational field of the earth is an example of a vector field describing the interaction between a massive object and the Earth. According to Newton’s universal law of gravitation, the gravitational force between two masses m and M is vector F_g = -G * Mm/r² * r-hat where r is the distance between them and r-hat is the cuit vector at position m pointing from M to m. The constant of proportionality is the gravitational constant, G, which is 6.67*10^-11N*m²/kg². The force is always attractive and its magnitude is proportional to the inverse square of the distance between masses. If M is the Earth’s mass, the gravitational field vector g at a point P in space is lim m to 0 of vector F_g/m -G(M/r²)r-hat. The field is radial and points toward the earth’s center. Near the surface, vector g is approximately g(vector) – g*r-hat, where g is approximately 9.8 m/s². A mass in a constant gravitational field only moves in the direction of the field when its initial velocity in the same direction.

The interaction between electric charges at rest is the electrostatic force. Electric charges can be positive or negative and electrostatic force between charges falls off as the inverse square of their distance of separation and can be attractive or repulsive. Given an object with charge Q, an object placed at point P and a distance r from Q will experience a Coulomb force, vector F_e=k_e(Qq/r²)r-hat, where h-hat is the unit vector from Q to q. The constant of proportionality, k_e, 9.0 * 10⁹N*m²/C² is the Coulomb constant. The electric field at P is vector E = lim q to 0 of vector F_e/q * k_e(Q/r²)r-hat. The SI unit of an electric field is newtons/coulomb or N/C. If Q is positive, the electric field points radially away from the charge. If it’s negative, the field points inward. The charge Q creates an electric field E(vector) which exerts force vector F_e of qE(vector) on q.

A magnetic field is another vector field and the most familiar source is a bar magnet, with a North pole and South pole for ends. Opposite poles attract each other while similar ones repel. A compass placed near a bar magnet will align with the direction of the magnetic field. This is because a magnetic compass has a tiny bar magnet rotating about a pivot point going through the center of the magnet. When placed near a bar magnet producing a magnetic field, it experiences a torque which tends to align the north pole of the compass with the external magnetic field. The earth’s magnetic field behaves similar to a bar magnet, though its south pole is in the northern hemisphere.

Vector fields would require more complex representations. One example is a vector field representation which plots arrows on a rectangular grid using orange positive charges and blue negative charges. Optionally, the length of the arrow may also correspond to its magnitude. Arrows point toward a negative charge which is the sink and away front the positive which is the source. A field line representation is one of the most common ways to draw a vector field. Field lines, or lines of force, are drawn starting at a point in space and moving a short distance in the direction of the vector field, then stopping and finding a new direction, repeating indefinitely. The line in space is always tangent to the local vector field and can give a good representation of the field’s properties. Field lines can’t cross each other as that would mean there are two intersecting fields. Lastly, the grass seeds or iron fillings representation is named after how the objects would align themselves in the field, with the long axis aligning with each local field direction. The second name is used for magnetic fields. These provide dense samplings of the field shapes. Since the field is a continuous feature of the space between charges, the area between field lines is more undrawn field lines. When field lines or grass seeds representations are animated, the direction of the pattern indicates the directory energy in an electromagnetic field flows in.

Chapter 2 – Coulomb’s Law

Benjamin Franklin termed charges positive and negative? A unit of charge is called a Coulomb (C). The smallest unit of “free” charge known is an electron or proton’s charge which has the magnitude of e=1.602×10^-19 C and a charge of any ordinary matter is quantized in integral multiples of e. An electron carries a unit of negative charge, -e and a proton a unit of positive charge, +e. In a closed system, the total amount of charge is conserved since charge can’t be created or destroyed, but can be transferred between bodies. In a system of two point charges, q₁ and q₂, separated by a distance r in vacuum, the force exerted by q₁ on q₂ is found with Coulomb’s law which states that vector F₁₂=k_e(q₁q₂/r²)r-hat, where k_e is the Coulomb constant and r-hat=r(vector)/r and is a unit vector directed from q₁ to q₂. Electric force is a vector and in SI units, the Coulomb constant k_e is 1/(4ℼ*ε₀) = 8.9875*10⁹N*m²/C² where ε₀=1/(4ℼ(8.99*10⁹N*m²/C²))=8.85*10^-12C²/N*m². This is known as the “permittivity of free space.” Force vector F₂₁=-F₁₂ with Newton’s third law. Gravitational effect can be neglected when dealing with electrostatic forces between particles.

Coulomb’s law applies to any pair of point charges and with more than two charges, the net force on one is the vector sum of forces exerted on it by the others. This is the superposition principle which says that F₃=F₁₃+F₂₃ (for a 3 charge system). Since forces aren’t in direct contact, one charge creates a field which acts on the other charge. An electric charge q creates an electric field that can be quantified using the force a positive test charge q₀ experiences at a point. The field, vector E=lim q₀ to 0 of vector F_e/q₀. The limit is taken to avoid disturbing the source charges. Charge q creates an electric field vector E which exerts force, vector F_e=q₀E(vector) on the test charge. By Coulomb’s law, the electric field at distance r from q is vector E = 1/(4ℼ*ε₀) * q/r² * r-hat. By the superposition principle, the total electric field due to a group of charges is the vector sum of individual charges, vector E=sum over i of vector E_i. In field line diagrams, field lines can show positive charge going outward and negative going inward by having the arrow point in different directions.

A pair of charges with equal magnitude and opposite signs is known as an electric dipole. If separated by a distance of 2a, the dipole moment vector p from -q to +q is p=2qa j-hat and the magnitude is 2qa, with q greater than zero. For a charge-neutral system with N charges, vector p=sum i=1 to N of q_ir_i(vector) where vector r_i is the position vector of charge q_i. HCL, CO, H₂0 and other polar modules are dipoles. When placing a dipole in an electric field, even though the net force vanishes, the field exerts a torque on the dipole and rotates the dipole clockwise so its moment (vector p) is aligned with the field. The cross product of the moment and electrical field is equal to the torque. The field does work to rotate the dipole. For angle dθ, this is dW=-τ dθ=-pE sinθ dθ. The torque opposes an increase to θ, and the total amount of work to rotate the dipole from angle θ₀ to θ is W=integral θ₀ to θ of (-pE sin θ)dθ=pE(cosθ-cosθ₀). Positive work is done when cosθ is larger and the change in pontiac energy, ΔU of the dipole is the negative work done, ΔU=U-U₀=-W=-pE(cosθ-cosθ₀) where U₀ = -PEcosθ₀, which is the potential energy at a reference point. Choosing a reference point of ℼ/2 so its potential energy is 0, in the presence of an external field, the electric dipole has potential energy, U=-pE cosθ=-p(vector)*E(vector). A system is at a stable equilibrium when its potential energy is a minimum, which takes place when dipole p is parallel to E, making U a minimum, U_min=-pE. When they are anti-parallel, U_max=+pE and the system is unstable. A non-uniform electric field would add a net force on the dipole in addition to the torque and the motion would combine linear acceleration and rotation. An example of this is a charged comb interacting with paper. The paper has “induced dipole moments” and the field on the comb is non-uniform due to its shape. Dipoles create electrical fields as well.

The superposition principle helps compute the electrical field for a small number of particles (discrete), but what about a very large number distributed in a region of space (continous). To find a field at point P, consider a small volume element, ΔV_i containing charge detlaq_i. The distances between charges in the volume are smaller than r, the distance between ΔV_i and P. As ΔV_i goes to zero, the volume charge density, ρ(vector r) = lim Δv_i to 0 of Δ_i/ΔV_i=dq/dV. ρ(r)’s dimension is charge/unit volume (C/m³) in SI units and the total charge in the volume V is Q=sum over i of Δ q_i=integral over V of ρ(vector r)dV. This is similar to mass density, ρ_m(r). When a large number atoms are tightly packed in a volume, the continuum limit and mass of an object is M=integral over V of ρ_m(vector r)dV.

Charge can be distributed across a surface S of area A with a surface charge density σ, σ(vector r)=dq/dA. σ’s dimension is charge/unit area (C/m²) in SI units. The total charge on the surface is Q=integral of integral over s of ρ(vector r)dA. (There is a double integral, there.) If charge is distributed over a line of length L, the linear charge density, lambda (C/m) is lambda(vector r)=dq/dL and the charge over the line is Q=integral over line of lambda(vector r)dL. If charges are uniformly distributed, then the densities are uniform. The electric field at P due to each charge element dq is dE(vector)=1/4ℼε₀ * dq/r² * r-hat where r is the distance from dq to P and r-hat is the corresponding unit vector. Via the superposition principle, the total electric field is the vector sum or integral of each field, E(vector) = 1/4ℼε₀ integral over V of dq/r² r-hat. This is a vector integral consisting of three integrations, one for each component of the field. At sufficiently far away from a continuous charge distribution of a finite extent, the electric field approaches its “point-charge” limit.

Session 3 Notes

The content of the first two session summaries in the course is all repeated information from the readings, but the third covers mathematical topics needed in the course. Maxwell’s equations involve line and surface integrals over open and closed surfaces. A closed surface has an inside and outside such as a basketball and an open surface has no inside or outside such as an infinitely flat thin plate. A two dimensional contour bounds an open surface (the plate’s rim) but not a closed surface. Maxwell’s equations for closed surfaces are double integral over S of E(vector)*dA(vector)=Q_in/ε₀ and double integral over S of B(vector)*dA(vector)=0. Maxwell’s equations for open surfaces are integral over C of E(vector)*d s(vector) = -d φ_B/dt and integral over C of B(vector) d s(vector)=mu₀I_enc+mu₀ε₀ dφ_E/dt. S is the surface and C is the contour bounding open surfaces. The line integral of a scalar function f(x,y,z) along path C is integral over C of f(x,y,z)ds=lim as N goes to infinity and Δ s_i goes to 0 of the summation from i=1 to N of f(x_i,y_i,z_i)Δ s_i where C is divided into N segments with length Δs_i. For a vector function, F(vector)=F_xi-hat+F_yj-hat+F_zk-hat, the line integral along path C is integral over C of F(vector)*d s(vector) = integral over C of F_xdx+F_ydy+F_zdz where d s(vector)=dx i-hat + dy j-hat+dz k-hat in the differential line element along C.

A function F(x,y) of two variables can be integrated over a surface S angiving a double integral. Double integral over S of F(x,y)dA = double integral over S of F(x,y)dx dy where dA=dx dy is a Cartesian differential area element on S. When F(x,y) is 1, A=double integral over S dA. For a vector function, vector F(x,y,z), the integral over a surface is double integral over S of F(vector)*dA(vector)=double integral over S of F(vector)*n-hat dA= double integral over S of F_ndA where dA(vector)=dA n-hat and n-hat is a unit vector pointing the direction of the surface. F_n is a dot product of F(vector) and n-hat and the component of F(vector) parallel to n-hat. The surface integral for a vector function is a flux, and for an electric field, E(vector) is less than the electric flux through a surface is φ_E=double integral over S of E(vector)*n-hat dA=double integral over S of E_ndA.

Chapter 3 – Electric Potential

Gravitation’s potential energy is a lot easier to visualize than electricity’s, and gravitational potential energy, U=mgh (in a uniform gravitational field g) only changes for a mass as the mass changes its position. To change the potential energy by ΔU, you’d need to do equal work pushing with a large enough external force that’s equal and opposite to gravity. If the force isn’t strong enough, gravity wins, and if the external force is too high, it will accelerate the object giving it velocity and kinetic energy.

Similar to gravitation, an electrostatic force, vector F_e has an inverse-square form that’s conservative in addition. In electric field vector E’s presence, the electric potential difference between points A and B is ΔV=-integral A to B of (vector F_e/q₀)*d s(vector)=-integral A to B of vector E *d vector s where q₀ is a test charge. The potential difference ΔV is the amount of work done per unit charge to move the test from point A to B without changing its kinetic energy. Electric potential isn’t the same as electric potential energy, which is ΔU=q₀ΔV. The SI unit of electric potential is volt V, where 1 volt is 1 joule/coulomb. At the atomic or molecular scale, a joule can be too large and electron volts (eV) are a more useful scale, which is the energy an electron acquires or loses when moving through the potential difference of one volt. 1eV=(1.6*10^-19C)(1V)=1.6*10^-19J.

Looking at a charge +q moving in the direction of a uniform electric field E(vector)=E₀(-j-hat), the path taken is parallel to vector E, so the potential difference between A and B is Δ V=V_B-V_A which ends up being greater than zero showing B has a lower potential. Electrical field lines always point from higher potential to lower and the change in potential energy is ΔU=U_B-U_A=-qE₀d. Since q is greater than zero, ΔU is greater than zero and the potential energy of a positive charge decreases as it moves along the direction of the electric field. This is similar to a mass losing potential energy as it moves in the direction of the gravitational field. If A to B is at an angle θ and not parallel to vector E, the potential difference becomes ΔV=V_B-V_A=-integral A to B of vector E*d s(vector)=-E(vector)*s(vector)=-E₀s cosθ=-E₀y. Moving along the electric field leads to a lower potential. The change in potential is the same regardless of the path taken since vector E is conservative. No work is done moving charges on the same equipotential line (with the same potential). This is shown by moving A to C to B instead to A to B, where C and B are the bottom of a right triangle.

If an electric field produced by charge +Q is vector E = (Q/4ℼε₀r²)r-hat, where r-hat points toward the field point, the potential difference between points A and B would be ΔV=V_B-V_A=-integral A to B of (Q/4ℼε₀r²)r-hat*d s(vector) and r-hat d s(vector) is ds cos θ = dr based on an image of the problem. The potential difference is only based on endpoints and not the taken path. Similar to gravity, only the difference in electrical potential is physically meaningful and a reference point with a potential of zero such as infinity can be chosen to help solve the problem. With a reference point of infinity, V_P=-integral infinity to P of E(vector)*d s(vector), the electric potential a distance “r” away from Q becomes V(r)=1/(4ℼε₀)(Q/r). With multiple charge points, the total electrical potential is the sum of potentials from individual charges by the superposition principle. A table shows a comparison of gravitation and electrostatistics. The main units are mass and charge respectively, a gravitational field’s equation is based on a gravitational force, while an electrical one takes Coulomb force. From there, potential energy charge and potential equations are the same based on their respective force or fields.

If a system of charges is assembled by an external agent, then ΔU = -W = +W_ext, meaning the change in the potential energy of the system is the work the agent puts in to assemble the configuration. For example, if you lift a mass, height h, you do +mgh work against a gravitational field’s -mgh. Charges are brought in from infinity without acceleration. For instance, if charges q₁ and P is r₁₂ and the goal is the work W₂ brings q₂ from infinity to P, W₂=q₂V₁, as no work is needed to set up the first charge. V₁=q₁/4ℼε₀r₁₂ and U₁₂ = W₂=(1/4ℼε₀)(q₁q₂/r₁₂). If the charges have the same sign, the work required is positive due to the repulsive force between the charges, otherwise it’s negative due to the attractive force between them. To add a third charge, q₃, W₃=q₃(V₁+V₂)=(q₃/4ℼε₀)((q₁/r₁₃)+(q₂/r₂₃)). The potential energy of the new configuration is U=W₂+W₃=U₁₂+U₁₃+U₂₃. The total potential energy in a system is the sum of contributions from distinct pairs, so for a system of N charges, U=(1/4ℼε₀) * sum i=1 to N of sum j=1 to N (where j > i) of (q_iq_j/r_ij). You can also count each pair twice and divide the result by two.

If the charge distribution is continuous, then the potential at P is the sum of all contributions from individual charges dq, such as V=1/4ℼε₀) * integral of dq/r, with infinity as a reference point with zero potential. Looking at two points, separated by a small distance d s(vector), dV=-E(vector) d*s(vector). For Cartesian coordinates E(vector)=E_xi-hat+E_yj-hat+E_zk-hat and d s(vector)=dxi-hat+dyj-hat+dzk-hat, then dV=(E_xi-hat+E_yj-hat+E_zk-hat)*(dxi-hat+dyj-hat+dzk-hat)=E_xdx+E_ydy+E_zdz, implying E_x=-∂V/∂x, E_y=-∂V/∂y, and E_z=-∂V/∂z. With a differential quantity called the del (gradient) operator, ∇=∂/∂x * i-hat + ∂/∂y * j-hat+ ∂/∂z k-hat and the electric field, vector E=E_xi-hat+E_yj-hat+E_zk-hat=-∇V. ∇ operates on a scalar quantity (electric potential) and gives a vector quantity (electric field). Mathematically, E(vector) can be viewed as the negative of the gradient of the electric potential V, though physically, the negative sign means if V increases as a positive charge moves in a direction, there’s a non-vanishing component of the field in the opposite direction. For gravity, if the gravitational potential increases when a mass is lifted, then the gravitational force is downward. If the charge distribution has spherical symmetry, the resulting field is a function of the radial distance, r, such as E(vector)=E_rr-hat. In this case, dV=-E_rdr. If V(r) is known, E(vector) = E_r r-hat=-(dV/dr)r-hat. If the potential due to a point charge q is V(r)=q/4ℼε₀r², the electric field is (q/4ℼε₀r²)r-hat.

If a system with two dimensions has an electric potential V(x,y), the curves characterized by a constant V(x,y) are equipotential curves. The same is true for V(x,y,z) for three dimensions. Since E(vector) = -∇V, the direction of E(vector) is always perpendicular to the equipotential through the point. In three dimensions, the potential change, dV is maximum when the gradient ∇V is parallel to d s(vector). This means ∇V always points in the direction of the maximum rate of change V with respect to displacement s. The properties of equipotential surfaces are as follows. The electric field lines are perpendicular to equipotentials and point from higher to lower potentials. By symmetry, the equipotential surfaces produced by a point charge form a family of concentric spheres and for a constant electric field, a family of planes perpendicular to the field lines. The tangential component of the field along the surface is zero and no work is required to move a particle along an equipotential surface. A topological map is a good visualization where each contour line is a fixed elevation above sea level.

Chapter 4 – Gauss’s Law

The strength of an electric field is proportional to the number of field lines per area. The number of field lines penetrating a given surface is an electric flux, denoted as φ_E. The electric field can be viewed as the number of lines per unit area. Given an area vector, vector A=A n-hat, with a magnitude of the surface area, A, and pointing in normal direction, n-hat, perpendicular to A, the flux through the surface is φ_E=E(vector)*A(vector)=E(vector)*n-hat A=EA. If the field, E(vector) makes an angle θ with n-hat, the electric flux becomes φ_E=E(vector)*A(vector)=EAcosθ=E_nA where E_n=E(vector)*n-hat is the component of E(vector) perpendicular to the surface. With the definition for the normal vector n-hat, the flux is positive if the field lines are leaving the surface and negative if entering.

A surface S can be curved and the field can vary over the surface. A surface is closed if it completely encloses a volume. To compute the flux, the surface is divided into a large number of infinitesimal area elements, ΔA(vector)_i=ΔA_in-hat_i. For a closed surface, n-hat_i points in the outward normal direction. In a case where an electric field passes through ΔA(vector)_i, making an angle θ with the normal of the surface, the flux through ΔA(vector)_i is Δφ_E*ΔA(vector)_i=E_iΔA_icosθ and the total flux of the surface is the sum over all the area elements. φ_E=lim as ΔA_i goes to 0 of sum of E_i(vector)*dA_i(vector)=double integral over S of E(vector)*d A(vector). Double integral is drawn as two integral signs with a ring through them and S is the closed surface. To calculate, specify the surface and sum over the dot product.

Given a positive point charge Q in the center of a sphere of radius r, the field due to the charge is E(vector)=(Q/4ℼε₀r²)r-hat, pointing in the radial direction. The charge is enclosed by an imaginary sphere of radius r called the “Gaussian surface.” In spherical coordinates, a small surface area element on the sphere is dA(vector)=r²sin θ d θ d φ r-hat. The shape of a Gaussian surface doesn’t have to be a sphere and will result in the same outward electric flux. This is due to Gauss’s law which states that the net flux through any closed surface is proportional to the net charge. This is also expressed as φ_E=double integral over S of E(vector)*d A(vector)=q_enc/ε₀ where q_enc is the net charge inside the surface. This works because the number of field lines that leave the charge is independent of the shape enclosing the charge.

A solid angle is used to prove Gauss’s law. If ΔA(vector)₁=ΔA₁ r-hat is an area element on the surface of sphere S₁ of radius r₁, the solid angle ΔOmega subtended by ΔA(vector)₁=ΔA₁ r-hat at the center of the sphere is ΔOmega=ΔA(vector)₁/r₁². Solid angles are dimensionless quantities measured in steradians (sr). Since the surface area of S₁ is 4ℼr₁², the total solid angle subtended by the sphere is omega=(4ℼr₁²/r₁)=4ℼ. A solid angle in three dimensions is analogous to the ordinary angle in two, where an angle Δ φ is the ratio of the length of the arc to the radius of a circle. If the arc’s length is s=2ℼr, the total angle is 2ℼ. A second area element in the three-dimensional example makes an angle θ with the radial unit vector r-hat and the solid angle subtended by ΔA₂ is ΔOmega=ΔA(vector)₂*r-hat/r₂²=ΔA₂cosθ/r₂²=ΔA_2n/r₂², where Δ A_2n = ΔA₂cosθ is the area of ΔA₂’s radial protection onto a second sphere S₂ of radius r₂, concentric with S₁. The subtended solid angle is the same for both ΔA₁ and ΔA_2n. If a point charge Q is placed at the center of the spheres, the electric fields strengths E₁ and E₂ at the center of area elements ΔA₁ and ΔA₂ are related by Coulomb’s law, E_i (1/4ℼε₀)(Q/r_i²) becomes E₂/E₁ = r₁²/r₂² and both have the same electric flux of E₁ΔA₁. The electric flux through any area element subtending the same solid angle is constant regardless of the shape or orientation of the surface.

Gauss’s law is useful for evaluating electrical fields, but is limited mostly to systems with cylindrical, planar, and spherical symmetry. In a cylindrical infinite rod system, the gaussian surface is a Coaxial Cylinder, for a planar infinite plane system, the Gaussian Pillbox is used and for a sphere or a spherical shell, the Gaussian surface is the Concentric Sphere. When applying Gauss’s law, identify the symmetry associated with the charge distribution, determine the electric field’s direction and a Gaussian surface on which the magnitude is constant over portions. Divide the space into regions associated with charge distribution and calculate q_enc, which is the charge enclosed by the Gassian surface, calculate flux throughout the surface for each region and equate φ_E with q_enc/ε₀ and deduce the magnitude of the electric field. A gaussian pillbox has a cylindrical shape and examples of solving all three of these are shown.

Insulators are materials such as glass or paper where electrons attach to atoms and can’t move freely. Conversely, in a conductor, electrons can move around freely. The electric field is zero in a conductor. For example, if a solid spherical conductor is placed in a constant external field, vector E₀, the positive and negative charges move toward the polar regions (left and right) of the sphere, inducing an electric field vector E’ which will point the opposite direction of the external field in the conductor. Charges will move until E’ completely cancels E₀. At electrostatic equilibrium, vector E vanishes in a conductor, while outside, E’ corresponds to a dipole field due to the induced charge distribution and the total electric field is E=E₀+E’. Any net charge in a conductor resides on the surface, as otherwise E would no longer be zero. The tangential component of E is zero on the surface. For an isolated conductor, the electric field is zero in the interior and distributes excess charge on the surface. The surface of a conductor in electrostatic equilibrium is an equipotential surface. Vector E is normal to the surface just outside the conductor. If its tangential component is initially non-zero, charges will move until it vanishes and only the normal component survives. In the examples of an infinitely large non-conducting plane and spherical shell, the normal component of the field exhibits a discontinuity at the boundary. A lightning rod’s design is based on the idea that the electric field strength on the surface of a conductor is the greatest at its sharpest point, shown in an example.

While at the boundary surface of a conductor with a uniform charge density σ, the tangential component of the electric field is zero and continuous, the normal component of the field has discontinuity with ΔE_n=σ/ε₀. Given a small patch of a conducting surface, the total force outside the surface is vector E = E(vector)_patch+E’(vector), with E’ being the charges not on the patch. By Newton’s third law, the patch cannot exert a force on itself and the force must come from E’. Assuming the patch is a flat surface, the electric field due to the patch is E(vector)_patch=+ σ/2ε₀ * k-hat when z is greater than 0 and -(σ/2ε₀)k-hat when z is less than 0. Due to the superposition principle, the field above the conducting surface is E_above=(σ/2ε₀)k-hat + E’(vector) and the field below is E_below=-(σ/2ε₀)k-hat + E’(vector). If E_above is (σ/ε₀)k-hat and E_below is 0, E(vector)_avg is σ/2ε₀ k-hat and the force acting on the patch is F(vector) = qE(vector)_avg= (σ A)*(σ/2ε₀)k-hat=(σ²A/2ε₀)k-hat, where A is the area of the patch. This is the force needed to drive charges on the surface of the conductor to an equilibrium state where the field outside the conductor takes on the value σ/ε₀ and vanishes inside. Electrostatic pressure on the patch is P=F/A=½ ε₀E² where E is the magnitude of the field just above the patch. The pressure is transmitted by the electric field.

Additional notes from Session 5 and 6

Gauss’s law is a new way of calculating electric fields from symmetric sources and is the first of Maxwell’s (four) equations. The idea behind the law is that electric fields flow in and out of charges and if you surround a region with a closed surface, you can see how much charge is enclosed by the region by seeing the fields flowing in and out. Surrounding a positive charge shows a net flow outward while a negative charge shows a net flow inwards. The magnetic Gauss’s Law is the second Maxwell equation, which is φ_B=double integral over S of B(vector)*d A(vector)=0, which also applies to closed surfaces. The remaining two maxwell equations concen open surfaces.

Electric fields superimpose and the field generated by a collection of charges is the sum of the fields generated by individual charges. If the charges are discrete then the sum is vector addition, if continuous, it requires integrating fields generated by small chunks of charge. Charge density is used to describe the amount of charge in a continuous charge distribution, telling thow much charge occupies a region of space at a point in time. Volume charge density is ρ=dq/dV, surface charge density is σ=dq/dA and linear charge density is lambda=dq/dl where V, A, and l stand for volume, area and length. The equation for a continuous charge distribution for point charge-like dq is E(vector)=1/(4ℼε₀*integral over V of (dq/r²) r-hat.

Chapter 5 – Capacitance and Dielectrics

A capacitor is a device which stores electric charge and a basic configuration is two conductors carrying equal but opposite charges. They are used in electronics to store electric potential energy, delay voltage changes (with help from resistors), filter unwanted frequency signals, form resonant circuits and make frequency dependent and independent voltage dividers (also when combined with resistors). In their uncharged state, a charge Q moves from one conductor to another giving one a charge +Q and the other -Q, creating potential difference ΔV, with the positively charged conductor having the higher potential. Whether charged or uncharged, the net charge of the whole capacitor is zero. A simple example of a capacitor shows two conducting plates of area A one above the other, separated by distance d. The amount of charge Q stored in a capacitor is linearly proportional to ΔV (the potential difference between plates), so Q=C|ΔV| where C is a positive proportionality constant called the capacitance. Physically, capacitance is a measure of the capacity of storing electric charge for ΔV, with a SI unit called a farad (F), where 1 F is 1 coulomb/volt. Capacitances are typically in the picofarad (1 pF=10^-12F) to millifarad range (1 mF=10^-3F=1000muF; 1 muF=10^-6F). In circuit diagrams, capacitors typically are shown with two parallel lines through the circuit line or one line and a curve for a polarized fixed capacitor with definite polarity.

If two metal plates with area A are separated by distance d, where the top has a charge +Q and the bottom has -Q, charging can be done by using a battery which produces a potential difference. To find the capacitance, the electric field between the plates is needed. A capacitor is finite in size, the field lines aren’t straight, and the field isn’t contained entirely between plates. This is known as edge effects and the non-uniform fields near the edge are fringing fields. For examples, field lines will be treated as straight lines even though they aren’t actually. To solve this, the limit is taken where plates are infinitely large and the field is calculated via Gauss’s law. A Gaussian pillbox with cap area A’ encloses the positive plates charge and the potential difference is found and the charge and magnitude of the potential difference are used to get the capacitance C, which is C=Q/|ΔV|. C only depends on the geometric factors A and d and increases linearly with A, since for a given potential difference ΔV, a bigger plate can hold more charge. C is inversely proportional to d, since the smaller the value of d, the smaller ΔV is for a fixed Q. Examples are shown with cylindrical and spherical capacitors where C still only depends on the physical dimensions. An isolated conductor (where a second conductor is placed at infinity) has a conductance and is calculated by setting the second as a reference point with a potential of zero.

A capacitor can be charged by connecting the plates to the terminals of a battery, which are maintained at a potential difference detlaV called the terminal voltage. Charges are shared between terminals and plates. The plate connected to the (positive) negative terminal will gain (positive) negative charge. This sharing causes a momentary reduction on charges in the terminals and decrease in terminal voltage. Chemical reactions are triggered to transfer more charge from one terminal to the other to compensate for the loss of charge to the capacitor plates and maintain the terminal voltage’s initial level. The battery is like a charge pump which brings charge Q from one plate to the other.

If two capacitors with different charges are connected in parallel, where left plates have are connected to the positive terminal of the battery and have the same electric potential and the same for the right plates and negative terminal, the potential difference, |ΔV| is the same across each capacitor, and each C_i is Q_i/|ΔV|. The two capacitors could be replaced with a capacitor with a total charge Q supplied by the battery, where Q=Q₁+Q₂ since Q is shared by the capacitors. The equivalent capacitance is C_eq=Q/|ΔV|=C₁+C₂. For N parallel capacitors, C_eq = sum i=1 to N of C_i.

For two initially uncharged capacitors connected in series, the potential difference |ΔV| is applied across both. In the example, the left place of the first capacitor is connected with the battery’s positive terminal and becomes positively charged with +Q. The right plate of the second is connected to the negative terminal and gets -Q charge as electrons flow in. The inner plates started uncharged but the outside plates attract an equal and opposite charge, so the first capacitor’s right plate gets -Q and the second’s left plate gets +Q. The total potential difference across any number of capacitors in series is the sum of the differences across the individual capacitors and the capacitors could be replaced with one equivalent capacitor, C_eq = Q/|ΔV|. |detlaV_i|=Q/C_i, and since potentials add in series, Q/C_eq=Q/C₁ + Q/C₂ and 1/C_eq=1/C₁ + 1/C₂. For N capacitors, 1/C_eq = sum i=1 to N of 1/C_i. In summary, the equivalent capacitance in parallel is C₁₂ = C₁ + C₂ while the equivalent capacitance in series is 1/C₁₂=1/C₁ + 1/C₂.

Capacitors can store energy and the amount of energy stored is equal to the work done to charge it. The battery works to remove charges from one plate and deposit them to the other. If the capacitor is uncharged initially and each plate has positive and negative charges, but they are balanced so there isn’t a net charge or electric field between the plates. The battery moves +dq charge from the bottom to the top plate, changing the positive charge to +dq and the bottom to -dq. By moving charge from the bottom to top plate, charge is built up on the capacitor and an electric field is generated. If the charge of the top plate is +q at some point and the potential difference between the plates is |ΔV|=q/C, to add charge +dq to the top plate requires dW=|ΔV|dq work to overcome electrical repulsion. If at the end of the process, the top plate is +Q, the total work done is W=integral 0 to Q of dq|ΔV|=integral 0 to Q of dq q/C=½ * Q²/C which is equal to the electrical potential energy, U_E of the system.

Energy stored in a capacitor can be viewed as being stored in an electric field itself. For a parallel plate capacitor with C=ε₀A/d and |ΔV|=Ed, potential energy, E_E=½*C|ΔV|²=½ * ε₀A/d * (Ed)² = ½ ε₀E²(Ad). Ad represents the volume between plates, so electric energy density u_E=U_E/Volume = ½ ε₀E². u_E is proportional to the square of the electric field. The energy stored in the capacitor can also be obtained from the point of view of external work as the plates are oppositely charged and force needs to be applied to maintain constant separation. If a small patch of charge Δ=ρ(ΔA) experiences an attractive force, ΔF=ρ²(ΔA)/2ε₀. If the total area of the plate is A, the external agent must exert F_ext=ρ²A/2ε₀ to pull the plates apart. The electric field strength in the region is E=ρ/ε₀, so the external force is F_ext=ε₀/2*E²A. The external force is independent of d and the work done to separate plates by distance d is W_ext=integral F(vector)_ext*d s(vector)=F_extd=((ε₀E²A/2)d. The potential energy of the system is equal to the work done by the external agent, so u_E=W_ext/Ad = ε₀E²/2. The electric energy density, u_E is also the electrostatic pressure P.

Many capacitors have an insulating material such as paper or plastic between the plates, as a dielectric used to maintain physical separation of the plates. Dielectrics break down less rapidly than air and can minimize charge leakage especially when high voltage is applied. Capacitance increases when the space between conductors is filled with dielectrics. For instance, given a capacitor with capacitance C₀ with no material between plates, when adding dielectric material to completely fill the space, capacitance increases to C=K_eC₀ where K_e is the dielectric constant. Different materials have different dielectric constants, though all are greater than 1. Every dielectric material has a characteristic dielectric strength of the maximum value of electric field before breakdown occurs and charges flow. Air has a constant of 1.00059 and strength in (10⁶V/m) of 3. Paper has a constant of 3.7 and strength of 16, glass has constant 4-6 and strength of 9 and water has constant of 80 and no strength. K_e is a measure of the dielectric response to an external electric field.

The two types of dielectrics are polar dielectrics and non-polar dielectrics. Polar dielectrics such as water have permanent electric dipole moments. The orientation of polar molecules is random without an electric field. When one is present a torque is set up and the molecules align with it. The alignment isn’t complete due to random thermal motion and the aligned molecules generate an electric field opposite the applied field but smaller in magnitude. Non-polar dielectrics don’t possess permanent dipole moments and electric dipole moments can be induced by putting the materials in an externally applied electric field. Surface charges produce an electric field in the opposite direction of the external field so the total field is a total of two fields, though the total has a lower magnitude than the external field.

For understanding dielectric materials, it’s important to know the average electric field produced by many small aligned dipoles. An example of this is a cylinder with area A and height h, consisting of N dipoles each with moment vector p spread uniformly through the volume of the cylinder. It’s assumed the moments are aligned with the axis of the cylinder. Each dipole has its own field, so what’s the average field ignoring external fields? To answer this, polarization vector P is the next dipole moment vector per unit volume. P(vector)=1/volume * sum i=1 to N of p(vector)_i. Since all dipoles are aligned, P=Np/Ah. Each positive charge of a dipole is cancelled on average by the negative charge of the dipole above it except for the dipoles at the top of the cylinder since there are no dipoles above them. The interior of the cylinder is uncharged in the average sense while the top surface has a net positive charge and bottom has a net negative charge. An expression for the equivalent charge Q_P is found by requiring the moment Q_P produces (Q_Ph) to be the total electric dipole moment of all of the dipoles, setting Q_Ph = Np and Q_P = Np/h. Computing the electric field produced by Q_P, the equivalent charge distribution resembles the parallel plate capacitor and the surface charge density ρ_P is equal to the magnitude of polarization, so ρ_P=Q_P/A=Np/Ah=P. P’s SI units are (C*m)/m³ or C/m², the same as surface charge density. If the polarization vector makes angle θ with n-hat (the outward normal vector of the surface), the surface charge density would be ρ_P=P(vector)*n-hat=Pcos θ. The equivalent charge system produces an average field of magnitude E_P=P/ε₀, with an opposite direction to P. E(vector)_P=-P(vector)/ε₀. The average field of the dipoles is opposite to the direction of the dipoles. If dipoles are randomly oriented and not aligned then the polarization would be zero and there wouldn’t be any average field. The non-vanishing average field appears when dipoles have a preferred orientation.

Looking at introducing dielectric material to a system, assume the molecules comprising the material have a permanent electric dipole moment. Since left to themselves dipoles won’t line up, in the absence of an external field P(vector)=0(vector) and E_P(vector) is zero as well. Placed in an external field E(vector)₀, a torque τ(vector)=p(vector)*E(vector)₀ aligns dipole vectors p(vector) with E(vector)_P, which is anti-parallel to E₀, which will tend to reduce the total electric field strength below E(vector)₀. The total electric field, E(vector)=E(vector)₀+E(vector)_P=E(vector)₀-P(vector)/ε₀. Generally, P is linearly proportional to E₀ and E since without the external field, there wouldn’t be alignment of dipoles or polarization. This linear relation is P(vector)=ε₀chi_eE(vector) where chi_e is the electric susceptibility. Materials that obey this relation are linear dielectrics. Combining equations, E₀=(1+chi_e)E(vector)=K_eE(vector) where K_e=(1+chi_e) is the dielectric constant and is always greater than 1 since chi_e is greater than zero, implying E=E₀/K_e is less than E₀. This shows dielectric materials always decrease the electric field. Without a battery, in the presence of a dielectric, an electric field decreases by a factor of K_e. The charge increases in the presence of a battery.

Gauss’s law can be used to find an electric field in the region between plates with no dielectric present, but when one is inserted, a charge Q_P is added in the opposite sign of the surface. Gauss’s law becomes double integral over S of E(vector)*d A(vector)=EA=(Q-Q_P)/ε₀. The dielectric weakens the original field E₀ by a factor of K_e and E=E₀/K_e=(Q-Q_P)/ε₀A. Q_P=Q(1-(1/K_e) and ρ_P=ρ(1-(1/K_e). In the limit as K_e, Q_P=0, corresponding to no electric material. Gauss’s law with dielectric can be written as double integral over S of E(vector)*d A(vector)=Q/K_eε₀=Q/ε, where ε=K_eε is the dielectric permittivity. Q = double integral over S of D(vector)*d A(vector) where D(vector)=ε₀K E(vector) is the electric displacement vector. Electromagnetic forces provide the glue for holding atoms together, keeping electrons near protons and binding atoms together in solids and an appendix goes into some detail on this.

Chapter 6 – Current and Resistance

Electric currents are flows of electric charge. The electric current is the rate charges flow across a cross-sectional area. If a collection of charges is moving perpendicular to a surface of area A, and ΔQ passes through the surface in time interval Δt, the average current is I_avg=ΔQ/Δt. The SI unit of current is ampere (A) with 1 A = 1 coulomb/sec. Common currents range from mega-amperes in lightning to nano-amperes in nerves. As Δt goes to zero, the instantaneous current I is dQ/dt. The direction of the current corresponds to the direction positive charges flow. Flowing charges in wires are negatively charged electrons moving in the opposite direction of the current. Currents flow in conductors (solids such as metals and semiconductors, liquids such as electrolytes and ionized ones, and ionized gasses) and flow is impeded in non-conductors and insulators.

Current is a macroscopic quantity and needs to be related to the microscopic motion of charges. Given a conductor with cross-sectional area A, the total current through the surface is I=double integral of J(vector)*d A(vector) where J(vector) is the current density in A/m² SI units. If q is the charge of each carrier and there are n charge carriers per unit volume, the total charge in the section is ΔQ=q(nA Δx). If carriers move with speed v_d, the displacement in Δt is Δx=v_dΔt and I_avg=ΔQ/Δt=nqv_dA. v_d is the drift speed, which is the average speed of the carriers inside a conductor when an external electric field is applied. An electron in the conductor doesn’t travel in a straight line and has an erratic path. The current density vector J is nq v(vector)_dd point in the same direction for positive carriers and opposite for negative. Electrons in the conductor experience electric force vector F_e = -e E(vector) giving acceleration vector a=F(vector)_e/m_e = -eE(vector)/m_e. The velocity of a given electron immediately after a collision is v_i and the velocity immediately before the next collision is v_f=v_i+at=v_i-eE/m_e * t where t is the time traveled. (v, a, and E are vectors). The average of v_f over all time intervals is v_f=v_i-(eE/m_e)t which equals the drift velocity v_d. Velocity of an electron is random without an electric field and v_i is 0. If τ = t is the mean free time between collisions, v_d=v_f=-eE/m_e * τ and the current density is J(vector)=ne²τ/m_e * E. Vectors J and E go in the same direction for either negative or positive carriers.

The current density for many materials is linearly dependent on the external electric field E, where J(vector)=σ E(vector). ρ is the conductivity of the material and that equation is the microscopic Ohm’s law. The equation works for ohmic materials and not for non-ohmic material. ρ = ne²τ/m_e. Looking at a segment of wire with length l and cross sectional area A, if the ends have a potential difference ΔV=V_b-V_a creating electric field E and current I, assuming E is uniform, ΔV=V_b-V_a=- integral a to b of E(vector)*d s(vector)=El. The current density is J=σE=σ(ΔV/l) and with J=I/A, the potential difference is ΔV=1/σ * J=(1/σA)I=RI where R=ΔV/I=l/σA is the resistance of the conductor. ΔV=IR is the macroscopic Ohm’s law and R’s SI unit the ohm represented by Omega, where 1Omega=1V/1A.

Most metals with good conductivity and low resistivity are ohmic. The resistivity ρ of a material is the reciprocal of conductivity. ρ=1/σ=m_e/ne²τ. ρ is related to resistance R of an object by p=E/J=RA/l or R=ρl/A. The resistivity of a material varies with temperature T and variation is linear for metals over a large T range. ρ=ρ₀[1+alpha(T-T₀)] where alpha is the temperature coefficient of resistivity. Typical values for ρ, σ and alpha for different materials are shown in a table comparing elements (such as iron, silver, and copper), alloys (brass, manganin and nichrome), semiconductors (carbon, germanium and silicon), and insulators (glass, sulfur and quartz), showing their resistivity, conductivity and temperature coefficient for 20 degrees celsius.

Looking at a circuit with a battery and resistor with resistance R, the potential difference between points a and b is ΔV=V_b-V_a which is greater than zero. If charge Δq is moved through the battery, the potential energy is increased by ΔU=ΔqΔV. As the charge moves around the resistor its potential energy is decreased due to collisions with atoms in the resistor. Neglecting the internal resistance of the battery and wires, returning to a, the potential of Δq is unchanged. The rate of energy loss through the resistor is P=ΔU/Δt=(Δq/Δt)ΔV=IΔV, which is the power supplied by the battery. Since ΔV=IR, P=I²R=(ΔV)²/R.

Chapter 7 – Direct-Current Circuits

Electrical circuits connect power supplies to loads such as resistors, motors, heaters or lamps. The connection between the supply and load is made by soldering with wires called leads or with connectors and terminals. Energy is delivered from the source to the user on demand with the flick of a switch. Many circuit elements can be connected to a common lead and various parts of the circuits are called circuit elements, which can be in series or parallel. Elements are in parallel when connected across the same potential difference. Loads are generally connected in parallel across the power supply. When elements are one after another, so the current passes through each without branches, the elements are in series. If there is a switch in a series, when the switch is open, the load is disconnected, when closed the load is connected. Currents flow when connected. Wires can touch and cause a short circuit. Most of the current will flow through the short and little through the lead possibly burning out electrical equipment. A fuse or circuit breaker is put in series to prevent damage, which will blow or open respectively on a short. In circuits, a point is chosen as the ground and assigned an arbitrary voltage (usually zero) with voltages at other points defined as the difference in voltage between that point and the ground.

Electrical energy must be supplied to maintain a constant current in a closed circuit. The energy source is the electromotive force or emf (ε), such as batteries, solar cells and thermocouples. These are charge pumps that move charges from a lower potential to the higher one. Emf ε = dW/dq, which is the work to move the unit charge. The SI for ε is volt (V). Given a circuit with a battery emf source and resistor with resistance R, if the battery has no internal resistance, the potential difference, ΔV (terminal voltage) between the positive and negative battery terminals is equal to the emf. To move the current, the battery discharges by converting chemical energy to emf. Since no work is done to move a charge around a closed loop due to the electrostatic force’s conservative nature, W=-q integral of E(vector) d s(vector)=0. If a is the starting point in the circuit (just before the negative terminal), when crossing from the negative to positive terminal, the potential increases by ε. After crossing the resistor, the potential decreases IR and potential energy converts to thermal energy in teh resistor. If the connecting wire has no resistance, the net change in the potential difference in the loop is zero, so ε – IR = 0 and thus I=ε/R. A real battery always has an internal resistance r and the potential difference across battery terminals becomes ΔV=ε-Ir. Since there’s no net change in potential difference in a closed loop, ε-Ir-IR=0 and I=ε/(R+r). The highest voltage occurs immediately after the battery dropping as resistors are crossed in the example. The voltage is essentially constant along wires which have negligibly small resistance compared to resistors. For a source with emf ε, the power or rate at which energy delivered is P=Iε=I(IR+Ir)=I²R+I²r. The power of the source emf is equal to the sum of the power dissipated in both the internal and load resistance is required by energy conversion.

Comparing resistors in series and parallel, in series, the total voltage drop across elements is the sum of the drop across individual resistors. ΔV=I(R₁+R₂) and two resistors can be replaced with an equivalent resistor for the same voltage drop, R_eq=R₁+R₂. Resistors in parallel cause the current to be divided to pass through each resistor which must satisfy Ohm’s law individually. So, ΔV_i=I_iR_i. The potential across resistors are teh same, where ΔV=ΔV_i=ΔV_i+1. Two resistors in parallel can be replaced with an equivalent with ΔV=IR_eq and thus 1/R_eq=1/R₁ + 1/R₂. All of these generalize to sums of N resistors. Most of the current will pass through the branch with the least resistance, so a short causes most current to pass through a path of near zero resistance.

Two fundamental (Kirchholff’s) rules are used to analyze circuits. The junction rule says at any point when there is a junction between various current carrying branches, by current conservation the sum of the currents into the node must equal the sum of currents out of the node to avoid charge buildup at the junction. In other words, sum I_in = sum I_out. The loop rule says the sum of voltage drops ΔV across circuit elements that form a closed circuit is zero.

Instruments that measure voltage or current impact the circuit they observe. Ammeter devices indicate the flow of the current via a display and will cause a voltage drop due to the ammeter’s resistance. An ideal ammeter has zero resistance but some may cause non negligible drops. Ammeters can be converted to voltmeters by putting a resistor in series with the coil movement. Putting them in parallel with the circuit element can allow the voltage to be found by measuring current and knowing ΔV=IR. The larger the resistance, the smaller the current diverted through a coil. An ideal voltmeter would have infinite resistance.

A capacitor in a circuit with current will charge at rate that decreases with time as the charge stored increases. In a charging capacitor, the charges and current as a function of time are q(t)=Q(1-e^t/RC) and I(t)=(ε/R)e^-t/RC. A charged capacitor in a closed circuit with no current will act as a voltage source driving current around the circuit. Electrons flow from its negative plate through the wire to its positive one and the voltage across the capacitor decreases as it loses strength. The equation describing this is q/C – IR =0. The current flowing away from the positive plate is proportional to the charge of the plate, I=-dq/dt. With time the voltage across the capacitor is V_C(t)=q(t)/C = (Q/C)e^{-t/RC and q(t)=Qe-t/RC.}

Chapter 8 – Introduction to Magnetic Fields

A charged object produces an electric field E at all points in space. Similarly, a bar magnet is a source of magnetic field B. Moving a compass near the magnet will cause the needle to line up with the field’s direction. The bar magnet consists of north and south poles where magnetic fields are strongest. Magnetic field lines leave the north and enerter the south poles. With two magnets, like poles will repel each other and opposites will attract. Electric charges can be isolated but magnetic poles always come in a pair. Breaking a bar magnet creates two new bar magnets each with north and south poles. An electric field E(vector) =F(vector)_e/q and is the force per unit charge. Since magnetic monopolies don’t exist, B can’t be defined the same way.

To define a magnetic field at a point, given a particle of charge q moving at velocity v, we know the magnitude of the magnetic force F(vector)_B exerted on the charged particle is proportional to v and q, the magnitude and direction of the force depend on v(vector) and B(vector). F_B disappears when v is parallel to B. When v makes angle θ with B, the direction of F_B is perpendicular to the plane formed by v and B and the magnitude of F_B is proportional to sin θ. When the sign of the charge of the particle changes, the direction of the magnetic force reverses. These observations are summarized as F(vector)_B=q v(vector)*B(vector). This is the working definition of the magnetic field at a pont in space and the magnitude of F(vector)_B=|q|vB sin θ. The SI unit of the magnetic field is the tesla (T) where 1 Tesla = 1 Newton/(Coulomb)(meter/second). The Gauss (G) is another commonly used non-SI unit and 1T=10⁴G. F(vector)_B is perpendicular to v(vector) and B(vector) and can’t change the particles speed v(not vector) or kinetic energy. Magnetic force can’t speed up or slow down a charged particle or do any work on the particle. The direction of v(vector) can be altered by the magnetic force.

A charged particle moving through a magnetic field experiences magnetic force F_B and an electric current or current-carrying wire in a magnetic field will also experience it. To calculate force on a wire, a segment of length l and area A is looked at. X’s are used to show the field points into the page. The charge moves at an average drift velocity v(vector)_d and total charge in the segment is Q_tot=q(nAl) where n is the number of charges per unit volume. The total magnetic force on the segment is F(vector)_B=Q_totv(vector)_d*B(vector)=qnAl(v(vector)_d*B(vector))=I(l(vector)xB(vector)) where I = nqv_dA and l(vector) is the length vector with magnitude l directed in the direction of the current. For a wire with an arbitrary shape, the magnetic force can be found by summing the forces on small segments denoted ds(vector). d F(vector)_B=Id s(vector)*B(vector) and the total force F(vector)_B=I integral a to b of d s(vector)*B(vector) where a and b are the endpoints of the wire. If a wire forms a closed loop of arbitrary shape, the force on the loop is F_B=I(integral of d s(vector))*B(vector). Since it’s a closed polygon and the vector sum is zero, the net magnetic force for a closed loop is zero.

Placing a rectangular loop carrying a current I in the xy plane and switching on a uniform magnetic field B(vector)=B i-hat running parallel to the loop causes sides that are parallel and anti-parallel to the field to vanish, since their cross products do. The sides of the loop perpendicular to the field lines are non-vanishing, though since they both go in different directions (in and out of page in the example), the net force on the rectangular loop is still zero. Even though the net force vanishes, remaining force vectors F₂ and F₄ produce a torque causing the loop to rotate on the y-axis. The torque with respect to the loop’s center is τ(vector)=(-b/2 * i-hat)*F(vector)₂+(b/2 * i-hat)*F(vector)₄=(-b/2 i-hat)*(IaB k-hat)+(b/2 i-hat)*(-IaB k-hat)=((IabB/2) + (IabB/2))j-hat=IabB j-hat = IAB j-hat. A=ab represents the loop’s area and the positive sign shows its clockwise about the y-axis. It’s convenient to introduce the area vector A = A n-hat with n-hat in the normal direction to the plane of the loop. This is set by the right hand rule and n-hat here is the positive k-hat direction. So, τ(vector)=IA(vector)*B(vector) and the magnitude of the torque is at its maximum when B is parallel to the plane or perpendicular to A. In a general situation where the loop of area vector A makes an angle θ with respect to the magnetic field, the lever arms can be expressed as r(vector)₂=b/2(-sin θ i-hat and cos θ k-hat)=-r(vector)₄ and the net torque becomes τ(vector)=IA(vector)*B(vector). For a loop with N turns, the magnitude of the toque is τ = NIABsin θ. NIA(vector) is the magnetic dipole moment, mu(vector), with a direction the same as A(vector), perpendicular to the plane of the loop and determined by the right-hand rule. The SI unit of mu is ampere-meter², A*m². The torque exerted on a current carrying loop can be written τ(vector)=mu(vector)*B(vector), analogous to τ(vector)=p(vector)*E(vector), the torque exerted on electric dipole moment p in the presence of electric field E.

The potential energy for a dipole is U=-p(vector)*E(vector) and the magnetic case is similar. The work done by an external agent to rotate a magnetic dipole from angle θ₀ to θ is W_ext=integral θ₀ to θ of τ dθ’ = ΔU = U-U₀. W_ext=-W again, where W is the work done by the magnetic field. If U₀ is zero and θ₀ is ℼ/2, the dipole in the presence of an electric field has potential energy of U=-muB cos θ=-mu(vector)*B(vector). Theres a stable equilibrium when vectors mu and B are parallel and U_min =-muB, while when they are anti-parallel U_max=+muB and the system is unstable.

The force experienced by a current carrying rectangular loop in a uniform magnetic field is zero, but if the field isn’t uniform a net force will act on the dipole. If small dipole vector mu is placed along the symmetric axis of a bar magnet, the dipole experiences an attractive force by the magnet (with non-uniform magnetic field) and external force is required to move the dipole right. The F_ext exerted by an agent to move the dipole Δx is F_extΔx=W_ext=ΔU=-muB(x+Δx)+muB(x)=-mu[B(x+Δx)-B(x)]. For a small Δx, F_ext=-mu dB/dx which is positive. F_B=mu dB/dx=d/dx (mu(vector)*B(vector)). Generally, the magnetic force experienced by dipole mu placed ina non-uniform magnetic field B is F(vector)_B=nabla(mu(vector)*B(vector)) where nambla = ∂/∂x i-hat + ∂/∂y j-hat + ∂/∂z k-hat is the gradient operator.

If a particle with a charge q and mass m enters a magnetic field of magnitude B with a velocity v(vector) perpendicular to the magnetic field lines, the radius of the circular path it follows is r=(mv/|q|B) and the angular speed of the particle is w=(|q|B)/m. In the presence of an electric field and magnetic field, the total force on a charged particle is F(vector)=q(E(vector)+v(vector)*B(vector)), known as the Lorentz force. By combining the two fields, particles which move with a certain velocity can be selected. This was used b J. J. Thomson to measure charge-to-mass ratio of electrons. In a device, electrons with charge q=-e and mass m are emitted. When electric and magnetic forces cancel, electrons can move in a straight path. The condition for cancellation is eE=evB, and v=E/B. Only particles with that speed will be able to move in a straight line. If the speed of electrons is v=sqrt((2eΔV/m)), then e/m=E²/2(ΔV)B² and by measuring E, ΔV and B, charge to mass ratio can be determined.

A mass spectrometer is a tool for measuring the mass of an atom, where a particle carrying a charge +q is sent through a velocity selector and fields satisfy E=vB, allowing chosen particles to move into a region where a second magnetic field B₀ pointing into the page is applied. The particle moves in a circular path with radius r and hits a photographic plate. r=mv/qB₀, and since v=E/B, m=(qB₀r/v)=qB₀Br/E.

Chapter 9 – Sources of Magnetic Fields

Currents which arise due to the motion of charges are the sources of magnetic fields. When charges move in a conducting wire and produce a current I, the magnetic field at any point P due to the current can be sound by adding up magnetic field contributions, dB(vector) from segments of wire ds(vector). The segments are a vector quantity with segment length for magnitude and point in the direction of the current flow. The infinitesimal current source is I d s(vector). If r is the distance from the current source to P and r-hat the corresponding unit vector, the Boit-Savart law says from the current source I d s(vector), dB(vector)=mu₀/4ℼ * (Ids(vector)*r-hat)/r² where mu₀ is a constant called the permeability of free space. Mu₀=4ℼ*10^-7T*(m/A). This is similar to Coulomb’s law for electric field due to a charge element dq, which is dE(vector)=1/4ℼε₀ * dq/r² * r-hat. Adding these contributions requires integrating over the current source, so B(vector)=integral over wire of dB(vector)=mu₀I/4ℼ * integral over wire of ds(vector)*r-hat/r². The integral is a vector integral so the expression for B is 3 integrals, one for each component of B. This nature is in the cross product Id s(vector) x r-hat and knowing how to evaluate the cross product and perform the integral is key for using the Biot-Savart law and there are detailed problem solutions showing this.

Given an infinitesimal current element in the form of a cylinder of cross sectional area A and length ds consisting of n charge carriers per unit volume, all moving at velocity v(vector) along the axis of the cylinder, I is the current in the element defined as the charge passing through any cross-section per unit time. I = nAq|v(vector)| and the total number of charge carriers in the current element is dN=nAds. The magnetic field dB from the dN charge carriers is dB(vector)=mu₀ * ((nAq|v(vector)|)d s(vector)*r-hat)/r² = mu₀/4ℼ * ((n A ds)qv(vector)*r-hat)/r²=mu₀/4ℼ * ((dN)q v(vector)*r-hat)/r² where r is the distance from the charge and teh field point P where the fiend is being measured. The unit vector r-hat = r(vector)/r points from the source of the field (the charge) to P. The differential length vector ds is parallel to v(vector)). For a charge dN=1, the last equation is B(vector)=mu₀/4ℼ * (qv(vector)*r-hat)/r². Since a point charge doesn’t constitute a steady current, the above equation only works in the non-relativistic limit where v is much smaller than c (speed of light) so the effect of retardation can be ignored. The same can work for N point charges. Using the superposition principle, the magnetic field at P is B(vector)=sum i=1 to N of mu₀/4ℼ * q_iv(vector)_i * [((x-x_i)i-hat+(y-y_i)j-hat+(z-z_i)k-hat)/[(x-x_i)²+(y-y_i)²+(z-z_i)²]^3/2].

Since a current-carrying wire produces a magnetic field and when placed in a magnetic field, a wire carrying a current experiences a net force, two current-carrying wires exert force on each other. Given two parallel wires separated by distance a and carrying currents I₁ and I₂ in the +x-direction, the magnetic force F(vector)₁₂ exerted on wire 1 by wire 2 can be computed. The magnetic field lines due to I₂ going in the +x-direction are circles concentric with wire 2, with the fiend B₂ pointing in the tangential direction. At an arbitrary point P on wire 1, B₂=-(mu₀I₂/2ℼa)j-hat, which points in the direction perpendicular to wire 1. F(vector)₁₂=I₁l(vector)*B(vector)₂=I₁(l i-hat)*(-mu₀I₂/2ℼa)j-hat = -mu₀I₁I₂l/2ℼa * k-hat. This points toward the second wire and two parallel wires carrying currents in the same direction attract each other while ones carrying currents in opposite directions repulse each other.

Currents are sources of magnetism. Without current, all compass needles point in the same direction. With current, needles are deflected along the tangential direction of a circular path in an example where a current is placed in the center of a circle of compasses. Dividing the circular path of radius r into many small length vectors Δs(vector)=Δs φ-hat, pointing the tangential direction with magnitude Δs. In the limit as Δs goes to zero, integral of B(vector)*d s(vector)=B integral of ds = mu₀I. This is obtained by choosing a closed path or “Amperian loop” following one particular magnetic field line. Considering a more complicated Amperian loop with two field lines, the line integral of the field around a contour abcda is integral over abcda of B(vector)*d s(vector)=integral over ab of B(vector)*d s(vector) + integral over bc of B(vector)*d s(vector) + integral over cd of B(vector)*d s(vector) + integral over cd of B(vector)*d s(vector) = 0+B₂(r₂θ)+0+B₁[r₁(2ℼ-θ)] where the length of arc bc is r₂θ and r₁(2ℼ-θ) for arc da. The first and third integrals disappear since the field is perpendicular to the paths of integration. If B₁=mu₀I/2ℼr₁ and B₂=mu₀I/2ℼr₂, the above becomes integral over abcda of B(vector)*d s(vector)=mu₀I/2ℼr₂ * (r₂θ)+(mu₀I/2ℼr₁ * [r₁(2ℼ-θ)]=mu₀I. The same result is obtained whether the closed path involves one or two field lines. The line integral of integral of B*ds around any closed Amperian loop is proportional to I_enc, the current encircled by the loop.

Generalizing to any closed loop regardless of shape, involving magnetic fields gives Ampere’s law, integral of B(vector)*d s(vector)=mu₀I_enc. This is to magnetism what Gauss’s law is to electrostatics. To apply Ampere’s law, the system must possess certain symmetry. For an infinite wire, the system has cylindrical symmetry and Ampere’s law works. If the length is finite, Biot-Savart law needs to be used instead. Ampere’s law works for infinitely long straight wires with steady current, infinitely large sheets of thickness and current density, an infinite solenoid and toroid. Examples of calculations for all four are shown.

A solenoid is a coil of wire wound in a helical form. If turns are closely spaced, the resulting magnetic field is fairly uniform, provided the length is much greater than its diameter. For an ideal solenoid which is infinitely long with tightly packed turns, the magnetic field inside is uniform and parallel to the axis, vanishing outside the solenoid. Ampere’s law can calculate the magnetic field strength here. Looking at a rectangular path of length l and width w, and traversing it counterclockwise, the line integral of B on the loop is integral of B(vector)*d s(vector)=integral over 1 of B(vector)*d s(vector)+integral over 2 of B(vector)*d s(vector)+integral over 3 of B(vector)*d s(vector)+integral over 4 of B(vector)*d s(vector)=0+0+Bl+0=Bl. Contributions from sides 2 and 4 are zero since B is perpendicular to ds and B is zero along side 1 as the magnetic field is only nonzero in the solenoid. The total current enclosed in the loop is I_enc=NI where N is the total number of turns. Applying Ampere’s law gives integral of B(vector)*d s(vector)=Bl=mu₀NI or B=mu₀NI/l=mu₀nI where n=N/l represents the number of turns per unit length. In terms of surface current or current per length unit, K=nI, the magnetic field can be written B=mu₀K.

If the length of the solenoid is finite, to find the field, you can approximate the solenoid as a large number of stacked loops. To find the field at point P on the z-axis, take a cross section of tightly packed loops at z’ with thickness dz’. The amount of current flowing through is proportional to the thickness of the cross-section and given by dI=I(ndz’)=I(N/l)dz’, where n=N/l is the number of turns per unit length. The contribution to the magnetic field at P due to the subset of loops is dB_z=(mu₀R²/2[(z-z’)²+R²]^3/2 (dIdz’). Integrating over the entire length, B_z=mu₀nIR²/2 * integral from -l/2 to l/2 of dz’/[(z-z’)²+R²]^3/2=mu₀nI/2 * [((l/2)-z)/sqrt((z-l/2)²+R²) + ((l/2)+z)/sqrt((z+l/2)²+R²)].

A Toroid is similar to a solenoid wrapped around with connected ends, like a wire wrapped around a donut. The magnetic field is completely confined and the field points in the azimuthal direction (clockwize). For a toroid consisting of N turns, using Ampere’s law, integral of B(vector)*d s(vector)=integral of Bds=B(2ℼr)=mu₀NI or B=mu₀NI/2ℼr, where r is the distance measured from the center of the toroid. Unlike a solenoid, a toroid’s magnetic field isn’t uniform and decreases as 1/r.

If a magnetic dipole moment mu(vector)=-mu k-hat is placed at an origin such as the center of the earth in the yz plane, what is the magnetic field at a point distance r from the origin. Y and z components of the magnetic field are B_y=-mu₀/4ℼ * (3mu/r³)sinθ cosθ, and B_z=-(mu₀/4ℼ)(mu/r³)(3cos²θ-1). In spherical coordinates, (r,θ,φ), the radial and polar components of the field are B_r=B_ysin θ+B_zcos θ=-(mu₀/4ℼ)(2mu/r³)cos θ and B_θ=B_ycos θ – B_z sin θ = -(mu₀/4ℼ)(mu/r³)sinθ. The magnetic field at the point due to the dipole is B(vector)=B_θθ-hat+B_rr-hat=-(mu₀/4ℼ)(mu/r³)(sin θ θ-hat+2cosθ r-hat). The negative sign is due to the dipole pointing in the -z direction and the magnetic field due to a dipole moment mu(vector) can be written B(vector)=(mu₀/4ℼ)((3(mu(vector)*r-hat)r-hat – mu(vector))/r³) and the ratio B_r/B_θ=2 cot θ.

The earth’s field acts as if there were a bar magnet in it. The south pole of the magnet is in the northern hemisphere to attract the north pole of a compass. The location of a point P on the surface is typically written in spherical coordinates (r, θ, φ), r distance from the center. θ is the polar angle from the z-axis, between zero and ℼ inclusive. φ is the azimuthal angle in the xy plane measured from the x-axis between zero and 2ℼ inclusive. If the distance is fixed at r=r_E (the radius of the earth), P is parameterized by two angles θ and φ. The typical numbers, latitude and longitude are related to θ and φ. Latitude, denoted Δ (lowercase) measures elevation from the plane of the equator, and is related to θ (the colatitude) by Δ = 90 degrees – θ. The equator has latitude zero degrees and the north and south poles are plus and minus 90 degrees respectively. The longitude of a location is φ in spherical coordinates and lines of constant longitude are meridians. The value depends on where the counting begins, historically passing through the Royal Astronomical Observatory in Greenwich, UK, which is chosen as the prime meridian with zero longitude. Letting the z-axis be the Earth’s rotation axis, and the x-axis passing through the prime meridian, the magnetic dipole moment of the Earth can be written mu(vector)_E=mu_E(sin θ₀ cos φ ₀ i-hat + sin θ₀ sin φ₀ j-hat + cos θ₀ k-hat) = mu_E(-0.062i-hat+0.18j-hat-0.98k-hat) where mu_E=7.79*10²²A*m², and (θ₀, φ₀) = (169, 109) in degrees. This shows mu_E has non-vanishing components in all three directions for cartesian coordinates. The location of MIT is 42N for latitude and 71W for longitude, so θ_m is 48 and φ_m is 289 and MIT’s position can be described with r(vector)_E(0.24i-hat-0.70j-hat+0.67k-hat) and the angle between -mu_E and r_MIT is θ_ME=cos^-1((-r(vector)_MIT*mu(vector)_E)/(|r(vector)_MIT||-mu(vector)_E|))=cos^-1(0.80)=37 degrees. The polar angle θ = cos^-1(r-hat*k-hat) which is the inverse of the cosine of the dot product between r-hat for the position and k-hat in the positive z-direction. The ratio of the radial to polar component of Earth’s magnetic field at MIT would be B_r/B_θ=2cot37=2.65. The positive radial/vertical direction points outward and positive polar/horizontal points to the equator here.

In electrostatics, dielectric materials always reduced the field below what it would be for a given amount of free electric charge. When dealing with magnetic materials, they can have multiple effects. Diamagnetic materials can reduce the magnetic field below what it would otherwise be for the same amount of current. Paramagnetic materials increase the magnetic field slightly above what it would otherwise be and ferromagnetic materials increase it a lot. Magnetic materials consist of many permanent or induced magnetic dipoles and the average magnetic field produced by many magnetic dipoles are aligned. An example shows a material in the form of a long cylinder with area A and length L, consisting of N magnetic dipoles each with a magnetic dipole moment mu spread uniformly through the volume of the cylinder and aligned with the axis of the cylinder. Each magnetic dipole has its own magnetic field and the magnetization vector M is the net dipole moment vector per unit volume, M(vector)=1/V * sum over i of mu_i where V is the volume and since all dipoles are aligned, the magnitude is M=N mu/AL. The small current loops are cancelled out by currents flowing in the opposite direction in neighboring loops and the only place without cancellation is near the edge of the cylinder where there aren’t adjacent loops further out. The average current in the interior of the cylinder vanishes and the sides have a net current. There is an equivalent current I_eq, which can be found by requiring the moment it produces to be the same as the moment of the system, so I_eqA=N mu and I_eq=(N mu)/A. The equivalent configuration is the same as a solenoid with a surface current K, and K=I_eq/L=(N mu)/AL=M. Surface current equals the magnetization, M, which is the average magnetic dipole moment per unit volume and the average magnetic field produced by the equivalent current system is B_M=mu₀K=mu₀M. Since the field has the same direction as M, B(vector)_M=mu₀M(vector). This is the opposite from electric dipoles where the field is anti-parallel from the electric dipoles. This relation is due to the average being taken and a different field will be observed looking at specific dipoles.

The atoms or molecules in paramagnetic materials have a permanent magnetic dipole moment. By themselves, the dipoles never line up spontaneously and without an external field, they are randomly aligned. The average field and the magnetism are both zero to start. If a paramagnetic material is placed in an external field B₀, the dipoles experience a torque τ(vector)=mu(vector)*B(vector)₀ that tends to align vectors mu and B₀ producing a net magnetization M parallel to B₀. Since B_M is parallel to B₀, it will tend to enhance B₀ and the total magnetic field B(vector)=B(vector)₀+B(vector)_M=B(vector)₀+mu₀M(vector). Despite both aligning parallel to the external field, the paramagnetic material enhances the external magnetic field while the dielectric material reduces an external electric field. Magnetization M is generally linearly proportional and in the same direction as B₀, which makes sense since alignment of dipoles and magnetization occur due to the fiend. The linear relation is M(vector)=chi_m (B(vector)₀/mu₀ where chi_m is a dimensionless quantity called the magnetic susceptibility. B(vector)=(1+chi_m)B(vector)₀=K_mB(vector)₀ where K_m=1 + chi_m is the relative permeability of the material, greater than 1 for paramagnetic material. The magnetic permeability of a material mu_m = (1+chi_mmu₀=K_mmu₀. In paramagnetic materials, mu_m is greater than mu₀.

For magnetic materials without permanent magnetic dipoles, the presence of external field B(vector)₀ induces dipole moments in atoms and molecules. These are anti-parallel to B₀ leading to anti parallel magnetization and average fields and a reduction in total magnetic field strength. For diamagnetic materials, the magnetic permeability is defined the same, though K_m is less than 1 or chi_m is less than zero and mu_m is less than mu₀. In ferromagnetic materials, there’s a strong interaction between neighboring atomic dipole moments. Ferromagnetic materials are made up of small patches called domains. An external field will line magnetic dipoles parallel to it and the strong interaction between dipole moments causes a much stronger alignment than in paramagnetic materials. The total magnetic field in a ferromagnet can be 10³ or 10⁴ times greater than applied field and the permeability K_m of a ferromagnetic material isn’t a constant since neither the total field or magnetization increases linearly with the external field. The relationship between B₀ and M isn’t unique and depends on the history of the material in a phenomenon known as hysteresis, making a hysteresis curve. In ferromagnets, the strong interaction between neighboring atomic dipole moments can keep them aligned even after the external field disappears and is the origin of permanent magnets.

Class 16 gives a good explanation of the right hand rules. For determining the direction of a dipole moment of a coil or wire, wrap your fingers in the direction of the current. Your thumb will point in the direction of the north pole of the dipole (in the direction of the moment of the coil). For determining the direction of the magnetic field generated by a current, fields wrap around currents in the same direction your fingers wrap around your thumb. The field is tangent to the circle your fingers make as you twist your hand keeping your thumb along the current. To determine the direction of force of a field on a moving charge or current, open your hand flat and put your thumb along v or I (for a current carrying wire) and fingers along B. Your palm points in the direction of the force. Torque is very similar to the second example, with the torque wanting to rotate the direction your fingers wrap around your thumb.

Chapter 10 – Faraday’s Law of Induction

Electric and magnetic fields seen so far have been produced by stationary and moving charges respectively. Imposing an electric field on a conductor leads to a current that generates a magnetic field. In 1831, Michael Faraday discovered that by varying magnetic field with time, an electric field could be generated. This idea is called electromagnetic induction. Faraday showed that no current registered in a galvanometer when a bar magnet was stationary with respect to the loop, but a current was induced in the loop when there was a relative motion between the magnet and loop. The galvanometer deflects in one direction as the magnet approaches and the opposite as it moves away. This experiment showed that an electric current was induced in the loop by changing the magnetic field. The coil acts as if it were conducted to an emf source, depending on the rate of change of magnetic flux through the coil.

Given a uniform magnetic field passing through a surface (S), if A(vector)=A n-hat, where A is the area and n-hat is the unit normal, the magnetic flux through the surface is φ_B=B(vector)*A(vector)=BAcosθ, where θ is the angle between B and n-hat. If the field isn’t uniform, φ_B=double integral over S of B(vector)*d A(vector). The SI unit of magnetic flux is the weber (Wb), where 1 Wb=1T*m². Faraday’s law of induction states that the induced emf ε in a coil is proportional to the negative of the rate of change in the magnetic flux, ε=-(dφ_B/dt). For a coil consisting of N loops, the total induced emf would be ε = -N(dφ_B/dt). For a spatially uniform field B(vector), ε=-(d/dt)(BAcosθ). An emf can be induced by varying the magnitude of vector B, A or the angle between them with time.

The direction of the induced current is determined with Lenz’s law which states that the induced current produces magnetic fields which tend to oppose the change in magnetic flux that induces such currents. For an example of this, given a conducting loop in a magnetic field, define a positive direction for vector A and assuming vector B is uniform, take the dot product of vectors B and A to determine the sign of the magnetic flux. Get the rate of flux change, dφ_B/dt by differentiating. The induced emf will have the same sign as the rate. Use the right-hand, with thumb pointing in A’s direction, curl your fingers around the closed loop. The induced current flows in the direction your fingers curl if emf is positive and the opposite direction if negative. Lenz’s law could be applied in a situation where a bar magnet moves toward a conducting loop with its north pole down. With the field pointing down area vector A pointing up, the magnetic flux is negative. As the magnet moves closer to the loop, the field at a point on the loop increases, producing more flux through the plane of the loop, leading to a positive induced emf and induced current flowing counterclockwise. The current sets up a magnetic field and produces a positive flux to counteract the change. Alternatively, the direction of the induced current can be determined from the point of view of magnetic force. Lenz’s law says the induced emf is in the direction opposing the change, so as the magnet approaches the loop, it experiences a repulsive force due to the induced emf. Like poles repel and the loop acts like a bar magnet with its north pole pointing up. Using the right-hand rule, the direction of the induced current is counterclockwise.

Given a conducting bar of length L moving through a uniform magnetic field pointing into pag, particles with charge q greater than zero experience a magnetic force F(vector)_B=qv(vector)*B(vector), tending to push them upward leaving negative charges on the lower end. The separation of charges causes an electric field E(vector) in the bar which produces a downward electric force F(vector)_e=qE(vector). Where the forces cancel, qvB=qE or E=vB and between the ends of the conductor, there’s a potential difference of V_ab=V_a-V_b=ε=El=Blv. ε aises from the motion of the conductor, so the potential difference is called the motional emf. The motional emf around a closed conducting loop can generally be written ε=integral(v(vector)*B(vector)*ds(vector) where ds(vector) is a differential length element. In the next example, the conducting bar moves through a region of uniform magnetic field B(vector)=-B k-hat, pointing into the page, by sliding along two frictionless conducting rails distance L apart and connected by a resistor with resistance R. An external force is applied so the conductor moves right with a velocity v in the i-hat direction and the magnetic flux through the closed loop formed by the bar and rails is φ_B=BA=BLx. According to Faraday’s law, the induced emf is ε = -dφ_B/dt = -BLv and the induced current I=|ε|/R=BLv/R in the counterclockwise direction by Lenz’s law. The magnetic force experienced by the bar as it moves right is F(vector)_B=I(L j-hat)*(-B k-hat)=-ILB i-hat=-(B²L²v/R)i-hat, in the opposite direction of v-hat. For the bar to move at a constant velocity, the net force acting on it must be zero and the external agent would have to supply a force of F(vector)_ext=-F(vector)_B, so the power delivered by F_ext is equal to the power dissipated in the resistor, P=F(vector)_ext*v(vector)=I²R, required by energy conservation. Without the force to keep a constant speed, the bar would slow down exponentially.

The electric potential difference between points A and B in an electric field E(vector) is ΔV=V_B-V_A=-integral A to B of E(vector)*d s(vector). When the electric field is conservative such as in electrostatistics, the line integral of E(vector)*d s(vector) is path independent and the integral of the closed path of E(vector)*d s(vector) is zero. Faraday’s law shows as a magnetic flux changes with time, an induced current begins to flow. The charges bove because of the induced emf which is the work done per unit charge. Since magnetic fields can’t do work, the work done on mobile charges is electric and the electric field in this case can’t be conservative as the line integral of a conservative fields would vanish. In this case, there’s a non conservative field associated with an induced emf, so ε=integral of E(vector)_nc*d s(vector) and via Faraday’s law, that equals -dφ_B/dt. This implies a changing magnetic flux will induce a non-conservative electric field that can vary with time. It’s important to distinguish between the induced, non-conservative electric field and conservative electric field arising from electric charges. For example, given a uniform magnetic field pointing into the page and confined to a circular region with radius R, if the magnitude of B(vector) increases with time, what’s the induced electric field everywhere due to this changing magnetic field. Since the magnetic field is confined to a circular region, symmetry can be used choosing an integration path that’s a circle of radius r. The magnitude of the induced field E(vector)_nc is the same at all points in the circle. By Lenz’s law, it’s direction needs to drive the induced current to produce a magnetic field opposing the change in magnetic flux. With area vector A pointing out of the page, the flux is negative or inward and since the magnitude of B increases with time, so does the flux. The induced current needs to flow counterclockwise to produce more outward flux.The rate of change of the magnetic flux where r is less than R, the rate of change of the magnetic flux is dφ_B/dt=d/dt(B(vector)*A(vector)=d/dt(-BA)=-(dB/dt)ℼr², which is equal to -integral of E(vector)_nc*d s(vector), implying E_nc=r/2 * dB/dt. For r greater than R, the induced electric field, E_nc(2ℼr)=-(dφ_B/dt)=(dB/dt)ℼR² or E_nc=R²/2r * dB/dt.

Generators and motors are important applications of Faraday’s law of induction. A generator converts mechanical energy into electric energy while a motor converts electrical energy into mechanical energy. A simple version of a generator shows an N-turn loop rotating in a uniform magnetic field. Magnetic flux varies with time, inducing an emf and flux φ_B=B(vector)*A(vector)=BAcosθ=BAcos ω t. The rate of change of the flux is dφ_B/dt=-BAω sin ω t. Since there are N turns in the loop, the total induced emf across the two ends is ε=-N dφ_B/dt=NBA ω sin ω t. Connecting the generator to a circuit with resistance R, the current generated by the circuit is I= |episolon|/R = NBAω/R * sin ω t. The current alternates and oscillates in sign with an amplitude I₀=NBAω/R. The power delivered to the circuit is P=I|ε|. The torque exerted on the loop is τ=muBsinθ=muBsin ω t and the mechanical power supplied to rotate the loop is P_m=τ ω. Since the dipole moment of the N-turn current loop is mu=NIA=(N²A²B ω)/R * sin ω t, P_m=(NAB ω)²/R * sin² ω t and the mechanical power put in equals the electrical power output.

When a conducting loop moves through a magnetic field, current is induced as the result of changing magnetic flux. If a solid conductor is used instead of a loop, an apparently circulating current called an eddy current is induced. This eddy also generates a magnetic force to oppose the motion and make it more difficult to move the conductor across the magnetic field. Since the conductor has non-vanishing resistance R, Joule heating causes power loss, P=ε²/R and by increasing R, power loss can be reduced. This could be by laminating the conducting slab or constructing the slab by gluing thin strips insulated from one another or cutting the slab to disrupt the conducting path. Eddy currents can help suppress unwanted mechanical oscillation and are used in the magnetic braking systems in high speed transit cars.

Chapter 11 – Inductance and Magnetic Energy

Given two coils placed near each other, where the first coil has N₁ turns and carries current I₁, creating a magnetic field B(vector)₁. Since the coils are close to each other, some of the field lines passing through coil 1 also pass through coil 2. φ₂₁ is the flux through 1 turn of the second coil due to I₁. By varying I₁ with time, there’s an induced emf associated with the changing magnetic flux in the second coil, ε₂₁=-N₂ * dφ₂₁/dt=-d/dt * double integral over coil 2 of B(vector)₁*dA(vector)₂. The time rate of change of magnetic flux φ₂₁ in coil 2 is proportional to the time rate of change in coil 1, N₂ * dφ₂₁/dt = M₂₁ * dI₁/dt and M₂₁ is a proportionality constant called the mutual inductance, also written as M₂₁=N₂φ₂₁/I₁. The SI unit for inductance is the henry (H) and 1H=1T*m²/A. Mutual inductance M₂₁ depends only on the geometrical properties of the coils such as the number of turns and radii of the two coils. If there were a current I₂ in the second coil, varying with time, the induced emf in coil 1 would be E₁₂=-N₁ * dφ₁₂/dt = d/dt * double integral over coil 1 of B(vector)₂*dA(vector)₁ and a current is induced in coil 1. The changing flux in coil 1 is proportional to the changing current in coil 2, N1 * dφ₁₂/dt = M₁₂ dI₂/dt where the proportionality constant M₁₂ is another mutual inductance that can be written M₁₂=N₁φ₁₂/I₂. With the reciprocity theorem which combines Ampere’s law and the Biot-Savart law, M₁₂=M₂₁=M.

Given a coil of N turns and carrying current I counterclockwise, if the current is steady, the magnetic flux remains constant. If the current changes with time, then according to Faraday’s law, an induced emf will arise to oppose the change. The induced current flows clockwise if dI/dt is positive and counterclockwise if negative. The property of a loop in which its own magnetic field opposes any change in current is called “self-inductance,” and the emf generated is called self-induced emf or back emf, denoted ε_L. All current carrying loops exhibit this property. An inductor is a circuit element with a large self inductance. E_L=-N * dφ_B/dt = -N d/dt * double integral of B(vector)*dA(vector) and is related to self-inductance L by E_L=-L dI/dt. Combining the equations, L=Nφ_B/I. Physically, the inductance of L is a measure of an inductor’s resistance to the change of current, where a larger L means a lower rate of change. L only depends on geometrical factors and is independent of the current.

Since an inductor in a circuit opposes changes in current through it, work must be done by an external source such as a battery to establish a current in the inductor. From the work-energy theorem, energy can be stored in an inductor. An inductor’s role is to magnetism what a capacitor is for electricity. The power or rate an external emf ε_ext works to overcome ε_L is P_L=dW_ext/dt = Iε_ext. If only the external emf and inductor are present, then the external emf and self induced are the same, implying P_L=dW_ext/dt = -Iε_L=+IL dI/dt. If the current is increasing, P is positive and the external source is doing positive work to transfer energy to the inductor, increasing the internal energy U_B of the inductor. If the current is decreasing, then P is less than zero and the external source removes energy from the inductor. The total work done by the external source to increase the current from zero to I is W_ext=integral of dW_ext=integral from 0 to I of LI’dI’=½ LI², which equals U_B, the stored magnetic energy in the inductor. U_B=½ LI² is similar to the electrical energy stored in a capacitor which is U_E=½ * Q²/C. From the energy perspective, there’s a distinction between an inductor and a resistor. When a current goes through a resistor, energy flows into the resistor and dissipates in the form of heat regardless of whether current is steady or time-dependent (since power dissipated in a resistor is P_R=IV_R=I²R). If energy flows into an ideal inductor only when the current varies with dI/dt greater than zero, the energy isn’t dissipated but is stored and then released when the current decreases with dI/dt < 0. If the current passing through the inductor is steady, there is no change in energy as P_L=LI(dI/dt)=0.

With time-changing magnetic fields added to simple circuits, the closed line integral of the electric field isn’t zero and for any open surface, the closed line integral of E(vector)*d s(vector)=-d/dt * double integral of B(vector)*d A(vector). Any circuit where the current changes with time has time-changing magnetic fields and induced electric fields. The electric potential difference between two points in the circuit is no longer path independent and the electric field isn’t conservative. The electric potential isn’t an appropriate concept as E(vector) can’t be written as the negative gradient of the scalar potential. Given a circuit, with a battery, a resistor, and a switch S that is closed at time zero and a one-loop inductor, the consequences of the inductance are shown. When time is greater than zero, current flows from the positive terminal of the battery to the negative. Applying Faraday’s law to the surface bounded by the current, dA is out of the page and ds is right handed (counter-clockwise). Finding the integral of the electric field around the circuit, there is an electric field in the battery and going from the positive to negative terminal while ds moves against it so E*ds is less than zero. The contribution of the battery to the integral is negative ε. There’s a field in the resistor in the direction of the current and when moving through it, E*ds is greater than zero, contributing +IR. Moving through the one loop inductor, there is no field if the resistance of the wire is zero, so closed line integral of E(vector)*d s(vector)=-ε + IR. The magnetic flux, φ_B through the open surface is positive as the current will produce a magnetic field B(vector) pointing out of the page (the same as the direction for dA). B is the self-magnetic field. From Faraday’s law, the line integral of E(vector) * d s(vector)=-ε+IR=- dφ_B/dt=-L dI/dt. The equation governing I(t) is ΔV=ε-IR-L(dI/dt)=0.

This is a modified Kirchhoff’s loop rule and the sum of the potential drops around a current is zero. To preserve the loop rule, the potential drop across an inductor needs to be specified. This rule can be obtained as the polarity of self-induced emf opposes the change in current by Lenz’s law. If the rate of change is positive, E_L creates an induced current in the opposite direction of I to oppose an increase. The inductor can be replaced with emf |ε_L|=L|dI/dt|=+L(dI/dt) less than zero. If there’s a decrease in current, the induced current set up by ε_L flows in the same direction as I to oppose the decrease. In both cases, the change in potential when moving from a to b along the direction of current I is V_b-V_a=-L(dI/dt). Kirchhoff’s loop rule modified for inductors says that if an inductor is traversed in the direction of the current, the potential change is -L(dI/dt). On the other hand, if the inductor is traversed in the direction opposite to the current, the potential change is +L(dI/dt). This will give correct answers for circuit problems, but its misleading and wrong sometimes in reality, as its based on the idea that the line integral of the electric field around a closed loop was zero which isn’t always true any more and the sum of potential drops won’t always be zero either and is +L(dI/dt).

A current won’t immediately rise to its maximum value due to the presence of a self induced emf in an inductor for RL circuits. An example circuit shows current from the positive end of an emf passing through a resistor, point A, inductor, point B, and a switch before going back to the negative terminal. An equivalent circuit replaces the inductor with an self-induced emf with its positive end facing the incoming current. Using the modified Kirchhoff’s rule for increasing current, ε – IR – |ε_L|=ε-IR-L(dI/dt)=0 for the given circuit. An important distinction between an inductor and resistor is that the potential difference across a resistor depends on I while it depends on dI/dt for an inductor. The self induced emf doesn’t oppose the current itself, but the change of current and the previous equation can be written dI/(I-ε/R)=-dt/(L/R). Integrating both sides and saying I(t=0)=0, I(t)=ε/R(1-e^-t/τ) where τ is L/R which is the time constant of the RL circuit. After a long enough time, the current reaches an equilibrium value ε/R. τ is a measure of how fast, the equilibrium is reached and the larger the value of L, the more time it takes. The magnitude of the self-induced emf is |E_L|=|-L(dI/dt)|=ε e^-t/τ which is at a maximum at time zero and vanishes as t reaches infinity. After long enough time, self-induction disappears and the inductor is just a conducting wire for two parts of a circuit. Energy is conserved in the circuit as I*ε=I²R+LI(dI/dt). I*ε is the rate the battery delivers energy to the circuit, I²R is dissipated power in the resistor and LI(dI/dt) is the rate energy is stored in the inductor. Whe energy dissipated is irrecoverable, magnetic energy stored in the inductor can be released later.

The next example looks at a circuit with decaying current, where the positive end of the field goes through switch 1 (starts closed) and then can take one of two loops back. The first loop goes through a second switch (which starts open) and the second goes through a resistor R and an inductor L. If a switch S₁ is closed long enough for current to be at equilibrium, what happens to the current when at time=0, the first switch is opened and the second is closed? An equivalent circuit shows current flowing through a self induced emf’s positive terminal and a resistor before going to E_L’s negative terminal. The current change is negative. Applying the modified Kirchhoff’s loop rule for decreasing current gives |ε_L|=-IR=-L(dI/dt)-IR=0 or dI/I=-dt/(L/R). I(t)=ε/R * e^-t/τ, which is the same time constant for the rising current.

The next circuit is an LC circuit showing a capacitor connected to an inductor. Current flows from the positive terminal of the capacitor with charge Q₀ to start, through a switch to an inductor and back to the negative terminal of the capacitor. When the switch is closed, the capacitor discharges and electric energy decreases. The current created from discharging generates magnetic energy which is stored in the inductor and without resistance, the total energy is transformed back and forth between electric energy in the capacitor and magnetic energy in the conductor, via electromagnetic oscillation. The total energy in the circuit after closing the switch is U=U_C+U_L=½ (Q²/C) + ½ LI². Since, U is constant, dU/dt = Q/C * dQ/dt + LI dI/dt = 0 and Q/C+L(d²Q/dt²)=0 where I = -dQ/dt and (dI/dt = -d²Q/dt². The general solution to this equation is Q(t)=Q₀(ω₀t+φ) where Q₀ is the amplitude of the charge and φ is the phase. Angular frequency ω₀ = 1/sqrt(LC) and the corresponding current in the inductor is I(t)=-dQ/dt=ω₀Q₀sin(ω₀t+φ)=I₀(ω₀t+φ) where I₀=ω₀Q₀. Since Q(t=0)=0 and I(t=0)=0, phase φ is zero and solutions for charge and current in the LC circuit are Q(t)=Q₀cos ω₀t and I(t)=I₀ sin ω₀t. At any instant of time, electric energy is U_E=Q²(t)/2C = (Q₀²/2C)cos²ω₀t and magnetic energy is U_B=½ LI²(t)=(Q₀²/2C)sin²ω₀t. Since total energy is the sum of the two, it remains constant, U=Q₀²/2C.

The mechanical analog to LC oscillations is the mass-spring system where a mass is attached to a wall by a spring. If the mass moves with speed v and the spring has spring constant k is displaced from equilibrium by x, the total energy of this mechanical system is U=K+U_sp=½ mv²+½ kx², where K and U_sp are kinetic energy of the mass and potential energy of the spring. Without friction, U is conserved and dU/dt=mv(dv/dt)+kx(dx/dt)=0 and the solution to the displacement is x(t)=x₀cos(ω₀t+φ) where ω₀=sqrt(k/m) is the angular frequency and x₀ is the amplitude of oscillations. At any point in time, the energy of the system is U=½ mx₀²ω₀²sin²(ω₀t+φ)+1/2 kx₀²cos²(ω₀t+φ)=½ kx₀².

An RLC circuit contains a resistor, inductor and capacitor. In an example, current flows from the positive end of a capacitor, through a switch (starting open) through an inductor and then a resistor. The capacitor starts with charge Q₀. Unlike the LC circuit, energy is dissipated through the resistor. Energy is dissipated at a rate dU/dt=-I²R, and Q/C*dQ/dt+LI * dI/dt=-I²R. Since current is equal to the rate of decrease of charge in capacitor plates, I=-dQ/dt and dividing both sides by I gives L(d²Q/dt²)+R(dQ/dt)+Q/C=0. For a small R, it’s easy to verify with Q(t)=Q₀e^-lambdatcos(ω’t+φ) where lambda=R/2L is the damping factor and ω’=sqrt(ω₀²-y²) is the angular frequency of damped oscillations. Constants Q₀ and φ are determined from initial conditions. In the limit where resistance vanishes, R=0 and the undamped, natural angular frequency ω₀=1/sqrt(LC) is recovered. The mechanical analog to the series RLC circuit is the damped harmonic oscillator.

When ω₀ is greater than lambda or ω’ is real and positive, a system is said to be underdamped. This occurs when the resistance is small. Charge oscillates(the cosine function) with exponentially decaying amplitude Q₀e^-lambdat. The frequency of a damped oscillation is less than that of an undamped oscillation where ω’ is less than ω₀. If ω₀ is less than lambda, w’ is imaginary and overdamping occurs. There is no oscillation in the overdamping case. The third damping case is critical damping, where ω₀=lambda, and ω’ is zero. There is no oscillation in this case either and energy decays rapidly. The quality of an underdamped oscillation is measured with quality factor, Q, where a larger Q means less damping and higher quality. Q=ω’(energy stored/average power dissipated)=ω’ U/|dU/dt|. The smaller the value of R, the greater the value of Q and the higher quality of oscillation.

The summary of class 21 has a section on transformers, which are an important use for mutual inductance and allow the modification of the voltage of AC signals. This takes an input signal and outputs a larger signal The primary coil takes an input and creates an oscillating magnetic field. An oscillating current is driven through it and the field is steered through an iron core (as ferromagnets like iron act like wires for magnetic fields), allowing the field to be bent in a loop. As the field goes through the turns of the secondary coil it generates an oscillating flux through them inducing an EMF in the secondary. If the core guides field lines well, the flux generated and received is proportional to the number of turns in each coil and the ratio of output to input is the ratio of the number of turns in the secondary to primary. With more turns in the secondary, a higher output is produced. Transformers are one of the main reasons AC power is used as opposed to DC. Before sending power across transmission lines, voltage is stepped up to a high voltage leading to lower energy loss as currents flow through the lines. It’s then stepped down to go into buildings.

Chapter 12 – Alternating-Current Circuits

Changing magnetic flux can induce an emf according to Faraday’s law of induction. If a coil rotates in the presence of a magnetic field, the induced emf varies sinusoidally with time and leads to an alternating current or AC. This provides a source of AC power. An example of an AC source is V(t)=V₀sin ωt where maximum value, V₀ is called the amplitude. Voltage varies between the positive and negative V₀. With the sine function, voltage at time t will be the same at time t’=t+T where T is the period. The frequency, f, is 1/T and has the unit of inverse seconds or hertz (Hz). The angular frequency is ω=2ℼf. If a voltage source is connected to an RLC circuit, energy is provided to compensate for energy dissipation in the resistor and oscillation won’t damp out. The oscillations of change, current and potential difference are called driven or forced oscillations. After an initial transient time, an AC flows in the circuit as a response to the driving voltage source. Current I(t)=I₀(ωt-φ) oscillates with the same frequency as the voltage source with amplitude I₀ and phase φ depending on the driving frequency.

A simple AC circuit connects a single circuit element to a sinusoidal voltage source, where there’s infinite capacitance and zero inductance. If the element is a resistor, then V(t)-V_R(t)=0. V_R(t) = I_RR is the instantaneous voltage drop across the resistor. The instantaneous current in the resistor is I_R(t)=V_R(t)/R=(V_R0sin ωt)/R=I_R)sin ωt. V_R0 is V₀ and I_R0=V_R0/R is the maximum current. φ is zero, so I_R(t) and V_R(t) are in phase with each other, reaching minimum or maximum values at the same time. This behavior is shown in a phasor diagram, where a phasor is a rotating vector where the length corresponds to the amplitude, rotates at angular speed ω, and the projection along the vertical axis corresponds to the value of the alternating quantity at time t. The phasor V_R0 has a constant magnitude of V_R0 and its projection in the vertical direction is V_R0ω t which equals V_R(t), the voltage drop across the resistor at time t. There’s a similar interpretation for I_R0, the current passing through the resistor. The diagram shows the current and voltage in phase with each other.

The average current over a period is [I_R(t)]=1/T integral 0 to T of I_R(t)dt = 1/T integral 0 to T of I_R0sim ω t dt = I_R0/T integral 0 to T of sin 2ℼt/T dt = 0. The average vanishes since [sin ωt]=1/T integral 0 to T of sin ω t dt = 0. Similar relations exist for other trig functions multiplied by ωt. The average of the square of the current is non vanishing due to this. [I_R²(t)]=½ I²_R0. The root-mean-square current is I_rms=sqrt([I_R²(t)]) and rms voltage V_rms=sqrt([V_R²(t)]). The rms voltage supplied to the domestic wall outlets is 120 V at frequency f=60 Hz in the USA. The power dissipated in the resistor is P_R(t)=I_R(t)V_R(t)=I_R²(t)R and the average over 1 period is [P_R(t)] = V_rms²/R.

Next, given a circuit with an AC generator and an inductor, this will correspond to infinite capacitance and zero resistance. Kirchoff’s rule for inductors gives V(t)-V_L(t)=V(t)-L(dI_L/dt)=0, implying dI_L/dt=V(t)/L=V_L0/L * sin ωt where V_L0=V₀. Integrating over this, I_L(t)=(V_L0/ωL)sin(ωt-(ℼ/2)) via the trig identity, -cos ω t = sin(ω t – ℼ/2). The amplitude of the current through the inductor is I_L0=V_L0/ωL=V_L0/X_L where X_L=ωL is the inductive reactance with SI units of ohms like resistance, though X_L depends linearly on angular frequency ω.Resistance to current flow increases with frequency as hat higher frequencies the current changes more rapidly. The inductive reactance vanishes as ω goes to zero. The phase constant is φ=+ ℼ/2. Current I_L(t) is out of phase with V_L(t) by φ and reaches its maximum a quarter of a cycle after voltage. This is stated as “the current lags voltage by ℼ/2 in a purely inductive circuit.

For a circuit with only an AC generator and a capacitor, both resistance and inductance are zero. Kirchoff’s voltage rule implies V(t)-V_C(t)=V(t)-Q(t)/C=0 and Q(t)=CV(t)=CV_C(t)=CV_C0sin ω t where V_C0=V₀. The current is I_C(t)=+dQ/dt=ωCV_C0sin(ωt+ℼ/2). The maximum value is I_C0=ωCV_C0=V_C0/S_C where X_C=1/ωC is the capacitance reactance with SI units ohms and representing effective resistance for a purely capacitive circuit. X_C is inversely proportional to C and ω and diverges as ω approaches zero. The phase constant φ is -ℼ/2. At time zero, voltage across the capacitor is zero while the current in the circuit is at its maximum. The current reaches its maximum before voltage by a quarter of a cycle and “the current leads voltage by ℼ/2 in a capacitive circuit.

Looking at a driven RLC circuit with an AC generator, capacitor, inductor and resistor, via Kirchhoff’s loop rule, V(t)-V_R(t)-V_L(t)-V_C(t)=V(t)-IR-L(dI/dt)-Q/C=0 giving the differential equation L(dI/dt)+IR+Q/C=V₀ sin ω. If the capacitor is uncharged initially, so I=+dQ/dt is proportional to its charge increase, the above equation can be written L(d²Q/dt²)+R(dQ/dt)+Q/C=V₀sin ωt and a possible solution is Q(t)=Q₀ cos(ωt-φ). Amplitude is Q₀=(V₀/L)/sqrt((Rω/L)²+(ω²-1/LC)²)=V₀/(ω*sqrt(R²+(X_L-X_C)². Phase is tan φ = 1/R(ωL-(1/ωC))=(X_L-X_C)/R and the circuit is I(t)=+dQ/dt=I₀sin(ωt-φ) with amplitude I₀=-Q₀ω. The current has the same amplitude and phase at all points in series RLC circuits and the instantaneous voltage across each of the three circuit elements has a different amplitude and phase relationship with the current. Instantaneous voltages are V_R(t)=I₀R sin ω t = V_R0sin ω t, V_L(t)=I₀X_L(t)=I₀X_Lsin(ωt+ℼ/2)=V_L0cos ω t, and V_C(t)=I₀X_C(ω t – ℼ/2)=-V_C0cos ωt where V_R0=I₀R, V_L0=I₀X_L, and V_C0=I₀X_C are the amplitudes of voltages across circuit elements The sum of all three voltages is equal to the instantaneous voltage from the AC source. V(t)=V_R(t)+V_L(t)+V_C(t). Or, V(phaser)₀=V(phaser)_R0+V(phaser)_L0+V(phaser)_C0. Current phasor I₀ leads the capacitive voltage phasor V_C0 by ℼ/2 and lags inductive voltage phasor V_L0 by ℼ/2. Voltage phasors rotate counterclockwise as time passes with fixed relative positions. Since voltages aren’t in phase with each other and reach maxima at different times, the maximum AC amplitude isn’t the sum of the circuit element’s maximum voltages.

The inductive reactance X_L=ωL and capacitance reactance, X_C=1/ωC are effective resistance in purely inductive and capacitive circuits, respectively. In a series RLC circuit, the effective is the impedance, Z=sqrt(R²+(X_L-X_C)², which has SI units of ohms. In terms of Z, current I(t)=V₀/Z * sin(ωt-φ). The amplitude of the current I₀=V₀/Z reaches a maximum when Z is at a minimum, which is when X_L=X_C, or ωL=1/ωC, leading to ω₀=1/sqrt(LC). The phenomenon at which current I₀ reaches its maximum is called a resonance and the frequency ω₀ is called the resonant frequency. At resonance, the impedance becomes Z=R, the amplitude of the current is I₀=V₀/R and the phase φ=0. In the series RLC circuit, instantaneous power delivered by the AC generator is given by P(t)=I(t)V(t)=V₀/Z * sin(ωt-φ)*V₀sin ω t = V₀²/Z * sin(ωt-φ)sin ω t = (V₀²/Z) (sin² ωt cos φ – sin ω t cos ω t sin φ) via the trigonometric identity sin(ω t – φ) = sin ω t cos φ – cos ω t sin φ. The time average of the power is [P(t)]=½ V₀²/Z * cos φ. In terms of rms quantities, the average power can be written [P(t)}=V_rms²/Z * cos φ = I_rmsV_rms cos φ. Here, cos φ is called the power factor and equals R/Z. At resonance, [P]_max=I_rmsV_rms=V_rms²/R.

A transformer is a device which can increase or decrease the AC voltage in a circuit and typically consists of two coils of wire, a primary and secondary, wound around an iron core. The primary coil has N₁ turns and is connected to alternating voltage source V₁(t). The secondary has N₂ turns and is connected to a load resistance R₂. Transformers operate based on the principle that an alternating current in the primary coil induces an alternating emf on the secondary due to their mutual inductance. In the primary, neglecting the small resistance of the coil, Faraday’s law of induction implies V₁=-N₁ (dφ_B/dt) where φ_B is the magnetic flux through one turn of the primary coil. The iron core extends from the primary to secondary and increases the magnetic field produced by the current in the primary coil and ensures that nearly all magnetic flux through the primary coil also passes through each turn of the secondary. The voltage or induced emf across the secondary coil is V₂=-N₂ * dℼh_B/dt.

In an ideal transformer, power loss due to Joule heating can be ignored and the power supply is completely transferred from the first to the secondary coil, so I₁V₁=I₂V₂. Also, no magnetic flux leaks out of the iron core and the flux through each turn is the same in both coils. Combining these gives the transformer equation V₂/V₁=N₂/N₁ and I₁=(V₂/V₁)I₂=(N₂/N₁)I₂. This means if the number of turns in the second coil is greater than the primary, the output voltage will also be greater than the input. If the secondary has less turns, then the output will be less. These are called step-up and step-down transformers, respectively.

The next example shows a parallel RLC circuit, where the current splits into a different path for each element. In the example, the AC voltage source is V(t)=V₀sin ω t. In parallel RLC circuits, the instantaneous voltages across all three circuit elements are the same and each voltage is in phase with the current through the resistor, though the currents through each element are different. The current in the resistor is I_R(t)=V(t)/R=(V₀/R) sin ω t = I_R0 sin ωt where I_R0=V₀/R. The voltage across the inductor is V_L(t)=V(t)=V₀sin ω t = L dI_L/dt and I_L(t)=integral 0 to t of V₀/L * sin ω t’dt’= I_L0sin(ω t – ℼ/2) where I_L0=V₀/X_L and X_L=ωL is the inductive reactance. Similarly, the voltage across the capacitor is V_C(t)=V₀sin ω t = Q(t)/C and I_c(t)=dQ/dt=I_C0sin(ω t + ℼ/2) where I_C0=V₀/X_C and X_C=1/ωC is the capacitive reactance. With Kirchhoff’s junction rule, the total current in the circuit is the sum of the three currents. I(t)=I_R(t)+I_L(t)+I_C(t)=I_R0sin ωt+I_L0sin(ω t – ℼ/2)+I_C0sin(ωt+ℼ/2). As phasers, I₀=I_R0+I_L0+I_C0 and the maximum amplitude of the total current, I₀ is the sum of phasers |I_R0+I_L0+I_C0|. Since I_R(t), I_L(t) and I_C(t) aren’t in phase, it isn’t equal to the maximum amplitudes of the three non phaser currents. As I₀=V₀/Z, the inverse impedance of the circuit is 1/Z=sqrt((1/R²)+(ωC-1/ωL)²). Phase is tan φ=R(ωC – 1/ωL) and the resonance condition for the parallel RLC circuit is given by φ=0 implying 1/X_C=1/X_L. The resonant frequency w₀=1/sqrt(LC) which is the same as for the series RLC circuit. 1/Z is minimum of Z is the maximum at resonance and the current in the inductor cancels the current in the capacitor, so the total current in the circuit reaches a minimum and is equal to the current in the resistor, I₀=V₀/R. Power is only dissipated through the resistor and average power [P(t)]=[I_R(t)V(t)]=[I_R²(t)R]=V₀²/2Z * (Z/R) with a power factor of [P(t)]/(V₀²/2Z)=Z/R=cos φ.

Chapter 13 – Maxwell’s Equations and Electromagnetic Waves

If a current-carrying wire possesses a certain symmetry, the field can be obtained by Ampere’s law, where the line integral of B(vector)*d s(vector)=mu₀I_enc, where B*ds is a magnetic field around an arbitrary closed loop and I_enc is the conduction current passing through the surface bound by the closed path. As a consequence to Faraday’s law of induction, a changing magnetic field can produce an electric field, line integral of E(vector)*d s(vector)=-d/dt * double integral over S of B(vector)*d A(vector). A time-varying electric field also produces a magnetic field. Looking at a capacitor being charged, during charging, the electric field strength increases with time as more charge accumulates on the plates. The conduction current that carries the charge also produces a magnetic field. To use Ampere’s law to calculate the field, choose curve a C to be the Amperian loop. If the surface bound by the path is a flat surface S₁, then the enclosed current is I_enc=I, but if a surface S₂ is chosen where no current passes through it, then I_enc=0. There’s an ambiguity in choosing the appropriate surface bounded by C. Maxwell shows the ambiguity can be resolved by adding a term to the right hand side of Ampere’s law, I_d=ε₀ * dφ_E/dt, which is the displacement current and involves a change in electric flux. The generalized Ampere’s law is now line integral of B(vector)*d s(vector)=mu₀I+mu₀ε₀ * dφ_E/dt = mu₀(I+I_d). In an example, the electric flux passing through S₂ is φ_E=double line integral over S of E(vector)*dA(vector)=EA=Q/ε₀ where A is the area of capacitor plates and I_d is related to the rate of charge increase on the plate, I_d=ε₀ * dφ_E/dt = dQ/dt, which is simply the conduction current, I, meaning the conduction current passing through S₁ is the same as the displacement current passing through S₂, so I=I_d. With the Ampere-Maxwell law, the ambiguity in choosing the surface bound by the Amperian loop is removed.

Gauss’s law for electrostatic states the electric flux through a closed surface is proportional to the charge enclosed. The electric field lines originate from the positive charge (source) and terminate at the negative charge (sink). Since no isolated magnetic monopole has been observed, Gauss’s law for magnetism is φ_B=double integral over S of B(vector)*dA(vector)=0, implying the number of magnetic field lines entering a closed surface is the same as the number leaving it. There is no source or sink and lines must be continuous without start or end points. For a bar magnet, field lines emanating from the north pole to the south pole outside the magnet return within the magnet forming a closed loop.

Maxwell’s equations are four equations which form the foundation of electromagnetic phenomena. Gauss’s law for E says electric flux through a closed surface is proportional to the charge enclosed. Faraday’s law says that changing magnetic flux produces an electric field. Gauss’s law for Bsays the total magnetic flux through a closed surface is zero, and the Ampere-Maxwell law says electric current and changing electric flux produce a magnetic field.

Looking at an electromagnetic wave propagating in the +x-direction with electric field E pointing in the +y-direction and magnetic field B in the +z-direction, this is an example of a plane wave since at any instant, E and B are uniform over any plane perpendicular to the direction of propagation. The wave is transverse since both fields are perpendicular to the direction of propagation, which points in the direction of cross product E(vector)XB(vector). With Maxwell’s equations, the relationship between the magnitudes of the fields can be found. Looking at a rectangular loop on the xy plane with the left side at x and right at x+Δx, the bottom side of the loop is at y and top is y+Δy. The unit vector normal to the loop is in the positive z-direction, n-hat=k-hat. With Faraday’s law, line integral of E(vector)*d s(vector)=-d/dt double line integral of B(vector)*d A(vector). The left hand side can be written integral of E(vector)*d s(vector)=E_y(x+Δx)Δy-E_y(x)Δy=[E_y(x+Δx)-E_y(x)]Δy=∂E_y/∂x * (ΔxΔy). The rate of change of magnetic flux on the right hand side is given by -d/dt * double integral of B(vector)*d A(vector)=-(∂B_z/Δt)(Δx Δy). Equating the two expressions and dividing through by the area ΔxΔy gives ∂E_y/∂x = -∂B_z/∂t. The second condition on the relationship between the fields can be deduced by the Ampere-Maxwell equation, line integral B(vector)*d s(vector)=mu₀ε₀ * d/dt * double integral of E(vector)*d A(vector). Given a rectangular loop in the xz plane with a unit normal n-hat=j-hat, the line integral of the magnetic field is line integral of B(vector)*d s(vector)=B_z(x)Δz-B_z(x+Δx)Δz=[B_z(x)-B_z(x+Δx)]Δz=-(∂B_z/∂x)(ΔxΔz) and the time derivative of the electric flux is mu₀ε₀*d/dt double line integral of E(vector)*d A(vector)=mu₀ε₀(∂E_y/∂t)(ΔxΔz). Equating the two and dividing by ΔxΔz, -∂B_z/∂x = mu₀ε₀(∂E_y/∂t). The result shows a time-varying electric field is generated by a spatially varying magnetic field. Both fields satisfy the one-dimensional wave equation, which can be shown by taking partial derivates with respect to x and t, ∂²E_y/∂x²=-∂/∂x * (∂B_z/∂t)=-∂/∂t * (∂B_z/∂x) = -∂/∂t ( (-mu₀ε₀ * ∂E_y/∂t)=mu₀ε₀ * ∂²E_y/∂t². Taking another partial derivative alternating respect to x and t gives ∂²B_z/∂x²=-∂/∂x * (mu₀ε₀ * ∂E_y/∂t)=mu₀ε₀ * ∂²B_z/∂t².

The general form of a one-dimensional wave equation is ((∂²/∂x²) – (1/v²) (∂²/∂t²)) ψ(x,t)=0 where v is the speed of propagation and ψ(x,t) is the wave function, E_y and B_z satisfy the wave equation and propagate with the speed of light. v=1/sqrt(u₀ε₀) = 1/sqrt((4ℼ*10^-7T*m/A)(8.85*10^-12C²/N*m²) = 2997*10⁸m/s = c, showing light is in an electromagnetic wave. It’s easy to verify any function of the form ψ(x + or – vt) satisfies the one-dimensional wave equation and a proof is shown using the chain rule. The wave equation is a linear differential equation meaning if phsi₁(x,t) and ψ₂(x,t) are solutions to the wave equation, then ψ₁(x,t) plus or minus ψ₂(x,t) is also a solution and electromagnetic waves obey the superposition principle. Example solutions to the wave equations are E(vector)=E_y(x,t)j-hat=E₀ cos k(x-vt)j-hat=E₀ cos(kx-ωt)j-hat and B(vector)=B_z(x,t)k-hat=B₀cos k(x-vt)k-hat=B₀cos(kx-ω t)k-hat where fields are sinusoidal, with amplitudes E₀ and B₀. The angular wave number k is related to wavelength lambda by k=2ℼ/lambda and angular frequency ω=kv=2ℼ * v/lambda=2ℼf where f is the linear frequency. In an empty space, the wave propagates at speed of light v=c. Fields E and B are always in phase, reaching minima and maxima together and E₀/B₀=ω/k=c. The fields magnitudes at any instant are related by E/B = c. The important features of electromagnetic waves in these equations are that the wave is transverse since both field vectors are perpendicular to the direction of propagation which points in the direction of the cross product of the fields. The fields are perpendicular to each other and their dot product vanishes. The ratio of the magnitudes and amplitudes of the fields is E/B=E₀=E₀/B₀=ω/k=c. The speed of propagation in vacuum is the speed of light and electromagnetic waves obey the superposition principle.

In a situation with two sinusoidal plane electromagnetic waves with one travelling in the +x-direction with E_1y(x,t)=E₁₀cos(k₁x-ω₁t) and B_1z(x,t)=B₁₀cos(k₁x-ω₁t) and the other wave traveling in the -x-direction with E_2y(x,t)=-E₂₀cos(k₂+ω₂t) and B_2z(x,t)=B₂₀cos(k₂x+ω₂t). For simplicity, the waves are assumed to have the same amplitudes and wavelengths, so all B’s, E’s, ω’s, and k’s are the same. With the superposition principle, the electric field can be written E_y(x,t)=E_1y(x,t)+E_2y(x,t)=E₀[cos(kx-ωt)-cos(kx+ωt)] and B_z(x,t)=B₀[cos(kx-ωt)+cos(kx+ωt)]. With the identities, cos(a+/-b)=cosacosb-/+sinasinb, E_y(x,t)=2E₀ sin kx sin ωt and B_z(x,t)=2B₀ cos kx cos ωt. These total fields still satisfy the wave equation and are standing waves which don’t propagate and oscillate in space and time. The electric field is zero at all times if sin kx is zero, or x=nℼ/k=nℼ/(2ℼ/lambda)=nlambda/2, where n=0,1,2, etc. Planes containing these points are nodal planes of the electric field, E(vector). Where kx is greater than 1, x=(n+½)(ℼ/k)=(n+½)(ℼ/(2ℼ/lambda))=((n/2)+(¼))lambda are anti-nodal planes of the electric field and the field’s amplitude is at its maximum 2E₀. In between two nodal planes, there is an anti-nodal plane and vice versa. For the magnetic field, nodal planes must contain points which meet the condition cos kx = 0, yielding x=(n+½)(ℼ/k)=((n/2)+(¼))lambda and anti-nodal planes for B(vector) contain points that satisfy cos kx = =/- 1 or x=(nℼ/k)=nℼ/(2ℼ/lambda)=nlambda/2. A nodal plane of E corresponds to an anti-nodal plane of B and vice versa. For time dependence, the electric field is zero everywhere sin ωt is zero, which is the maximum condition for the magnetic field. Unlike the traveling electromagnetic wave, where electric and magnetic fields are always in phase, in standing electromagnetic waves, the fields are 90 degrees out of phase. Standing waves can be formed by confining electromagnetic waves in two perfectly reflecting conductors.

Both electric and magnetic fields store energy, so energy can be carried by electromagnetic waves which consist of both fields. Given a plane electromagnetic wave passing through a small volume element of area A and thickness dx, the total energy in the volume is dU=uAdx=(u_E+u_B)A dx = ½(ε₀E²+ B²/mu₀)Adx where u_E=½ ε₀E² and u_B=B²/2mu₀ are the energy densities associated with the fields. Since the wave propagates with the speed of light, the amount of time it takes for the wave to move through the volume element is dt = dx/c and the rate of change of energy per unit area is S=dU/Adt=c/2 (ε₀E² + B²/mu₀). The SI unit of S is W/m². Since, E is cB and c=1/sqrt(mu₀ε₀), S=EB/mu₀. The rate of energy flow per unit area can be described by the Poynting vector S = 1/mu₀ * E(vector)XB(vector), pointing in the direction of propagation. Since the fields are perpendicular, the magnitude of S, |S(vector)|=|E(vector)XB(vector)|/mu₀ = EB/mu₀=S. For example, if the electric component of the wave is E(vector)=E₀cos(kx-ωt)j-hat, the corresponding magnetic component is B(vector)=B₀cos(kx-ωt)k-hat and the direction of propagation is +x. The Poynting vector is then S(vector)=1/mu₀(E₀cos(kx-ωt)j-hat)x(B₀(kx-ωt)k-hat)=E₀B₀/mu₀ * cos²(kx-ωt)i-hat. S(vector) points in the direction of wave propagation and the intensity of the wave, I is the time average of S, where I=[S]=E₀B₀/mu₀ [cos²(kx-ωt)]=E₀B₀/2mu₀=cB₀²/2mu₀ where [cos²(kx-ωt)]=½. Relating energy density to intensity, the electric and magnetic energy densities are equal to each other (u_B=u_E). The average total energy density [u]=[u_E+u_B]=B₀²/2mu₀ and intensity is related by average energy density by I=[S]=c[u]. An example of a Poynting vector at the upper surface of the Earth’s atmosphere, has a time-averaged magnitude called the solar constant [S]=1.35*10³ W/m² and the example walks through finding the magnitudes of the Sun’s electromagnetic radiation and time averaged-power.

Since the Poynting vector represents the rate of energy flow per unit area, the rate of change of energy in a system is dU/dt=- double line integral of S(vector)*dA(vector) where dA(vector)=dAn-hat, where n-hat is a unit vector in the outward normal direction. This treats S(vector)0 as the energy flux density, similar to the current density J(vector). If energy flows out of the system then S(vector)=S n-hat and dU/dt is less than zero, showing an overall decrease of energy in the system. If energy flows into the system, S(vector)=S(-n-hat) and dU/dt is greater than zero.

The electromagnetic wave transports both energy and momentum and can exert a radiation pressure on a surface due to the absorption and reflection of the momentum. Maxwell showed that if the plane electromagnetic wave is completely observed by a surface, the momentum transfer is related to the energy absorbed by Δp=ΔU/c in complete absorption. If the wave is instead reflected by a surface such as a mirror, Δp = 2ΔU/c in complete reflection. For complete absorption, the average radiation pressure (force per unit area) is P=[F]/A=1/A * [dp/dt] = 1/AC * [dU/dt]. Since the rate of energy delivered to the surface is [dU/dt]=[S]A=IA, P=I/c (complete absorption). If radiation had completely reflected, radiation pressure would be twice as great as complete absorption, P=2I/c.

Electromagnetic waves are produced when electric charges aceleartes, meaning a charge radiates energy when it undergoes acceleration. Radiation can’t be produced by stationary charges or steady currents. A common way of producing electromagnetic waves is to apply a sinusoidal voltage source to an antenna, causing charges to accumulate near the tips of the antenna. This produces an oscillating electric dipole. To make an electromagnetic plane wave, an electric field line acts similar to a string with an associated tension. Trying to displace it causes a restoring force and a wave propagates on the field line as a result.

Class 29 notes that changing magnetic fields creating electric fields and vice versa allows the fields to propagate through space in electromagnetic radiation, which is the single most useful discovery in the class helping to explain light and is used for radio, television and cell phone signals. EM waves travelling through space can carry light with them.

Class 32’s notes talk about polarization. EM waves are transverse waves and E and B fields are perpendicular to the direction of propagation p-hat as well as to each other. Given p-hat, the fields can oscillate along an infinite number of directions (any direction perpendicular to p-hat). The axis the E field is oscillating along is the polarization axis. A polarization direction is sometimes stated, but since the electric field oscillates, it sometimes points in that direction and sometimes points the opposite direction. When light has a specific polarization direction, it is polarized. Most light, such as from the sun or light bulbs is unpolarized, and electric fields oscillate along lots of different axes, but in certain cases light becomes polarized. For instance, if light scatters off a surface, only the polarization parallel to the surface survives. Polarized sunglasses stop all horizontally polarized light and block a large fraction of light which reflect off horizontal surfaces (glare). Rainbows are polarized and under some conditions the sky is, since the blue light in the sky is scattered sunlight.

Chapter 14 – Interference and Diffraction

Looking at a region in space where two or more waves pass through at the same time, according to the superposition principle, the net displacement is the vector or algebraic sum of the individual displacements. Interference is the combination of two or more waves to form a composite wave based on this. Given wavesψ₁(x,t)=ψ₁₀sin(k₁x +/- ω₁t+φ₁), andψ₂(x,t)=ψ₂₀sin(k₂x +/- ω₂t+φ₂), the resulting wave is the sum of the two waves. Interference is constructive if the amplitude of the resulting wave, ψ(x,t) is greater than the individual ones and destructive if smaller. For example, looking at the superposition of both waves at t=0, ψ₁(x)=sin x and ψ₂(x)=2sin(s+ℼ/4). The resulting wave ψ(x) is the sum of the two, (1+sqrt(2))sin x + sqrt(2) cos x thanks to the trigonometric identity, sin(alpha+beta)=sin alpha cos beta + cos alpha sin beta and sin(ℼ/4)=cos(ℼ/4)=sqrt(2)/2. To form an interference pattern, the incident light must satisfy two conditions. The light sources must be coherent, meaning the plane waves front eh sources must maintain a constant phase relation. If two waves are out of phase with φ=ℼ, this phase difference can’t change with time. The light also needs to be monochromatic, consisting of a single wavelength lambda=2ℼ/k. Light emitted from an incandescent lightbulb is incoherent because light consists of waves of different wavelengths without a constant phase relationship and no interference pattern is observed.

Thomas Young demonstrated the wave nature of light in 1801 with his double-slit experiment. A monochromatic light source is incident on a first screen with a single slit S₀ and the emerging light goes two a second screen with two parallel slits S₁ and S₂ serving as sources of coherent light. The waves form the two slits emerge and interfere to form an interference pattern on a viewing screen and bright bands or fringes correspond to interference maxima and dark to minima. Light falling on the screen at a point P a distance y from point O, lying on the screen a perpendicular distance L from the double slit system, where the slits are separated by distance d, will cause the light from slit 2 to travel an extra distance Δ=r₂-r₁ to point P than the light from slit one. The extra distance is the path difference. In the limit where the distance to the screen is much greater than the distance between slits, the path difference becomes d sin θ and the rays are basically parallel. Whether the waves are in phase or out of phase is determined by this value. Constructive interference occurs when Δ is zero or an integer multiple of the wavelength lambda. This multiple is called the order number the zeroth order (m=0) maximum corresponds to the central bright fringe at θ = 0. The first order maxima (m=+/-1) are the bright fringes on either side of the central fringe. When Δ is an odd integer multiple of lambda/2, the waves are 180 degrees out of phase at P and destructive interference occurs causing a dark fringe.

The total instantaneous electric field at point P is the vector sum of the two sources, while the Poynting flux S is proportional to the square of the total field, S proportional to E²=(E(vector)₁+E(vector)₂)²=E₁²+E₂²+2E(vector)₁*E(vector)₂. Taking the time average of S, the intensity I of the light at P is I=[S] proportional to [E₁²]+[E₂²]+2[E(vector)₁*E(vector)₂]. Cross term 2[E₁*E₂] represents the correlation between the two light waves. For incoherent sources, as there’s no definite phase relation between them, the cross term vanishes and the intensity due to the incoherent source is the sum of the individual intensities. For coherent sources, the cross term isn’t zero and for constructive interference where the fields are equal, the I=4I₁, four times greater than a single source’s intensity. When E₁=-E₂, destructive interference occurs and intensity becomes I=0. If the waves emerged from the slits are coherent sinusoidal plane waves and the field components are E₁=E₀sin ω t and E₂=E₀sin(ω t + φ), where the waves have the same amplitude of E₀. The point P is chosen as the origin so the kx dependence in the wave function is eliminated. SInce the wave from slit 2 traveled an extra distance, E has an extra phase shift φ relative to E₁ from slit 1. For constructive interference, a path difference of Δ=lambda corresponds to a phase shift of φ=2ℼ and Δ/lambda=φ/2ℼ or φ=(2ℼ/lambda) * Δ = (2ℼ/lambda) * d sin θ. If both fields point the same direction the total field via the superposition principle is E=E₁+E₂=E₀[sin ω t + sin(ω t + φ)] = 2E₀ cos(φ/2)sin(ω t + φ/2). Intensity I is proportional to the time average of the square of the total electric field. I proportional to [E²] and I=I₀cos²(φ/2) where I₀ is the maximum intensity on the screen.

In addition to interference, waves exhibit diffraction. Diffraction is the bending of waves as they pass by objects through an aperture. Huygen’s principle states that every unobstructed point on a wave front acts as a source of secondary spherical waves. The new wavefront is the surface tangent to all secondary spherical waves. This means that light waves incident on two slits will spread out and exhibit an interference pattern in the region beyond, called a diffraction pattern. If no bending occurs and waves continue to travel in straight lines, there would be no diffraction pattern. This chapter only looks at a special case of diffraction called Fraunhofer diffraction, where all light rays that emerge from the slit are approximately parallel to each other. For a diffraction pattern to appear on the screen, a convex lens is placed between the slit and screen to provide convergence of light rays.

So far, the width of slits was assumed to be small enough that each slit was a point source, but following examples with finite widths can show how Fraunhofer diffraction arises. A source of monochromatic light is incident on a slit of width “a” in the example. All rays passing through are roughly parallel and each portion of the slit acts as a source of light waves according to Huygen’s principle. Dividing the slit into two halves, at the first minimum, each fray front the upper is 180 degrees out of phase with a ray from the lower. For example, with 100 point sources, 1 to 50 are in the lower half and 51 to 100 are in the upper. Source 1 and 51 are separated by distance a/2 an out of phase with path difference Δ=lambda/2. A similar observation applies to 2 and 52 and any pair a distance a/2 apart. The condition for the first minimum is (a/2)sin θ = lambda/2 or sin θ = lambda/a. With the same logic for four equally spaced points a distance a/4 apart, the path difference would be Δ=a sin θ/4 and the condition for destructive interference is sin θ = 2lambda/a. This can be generalized to show destructive interference occurs when a sin θ = m*lambda where m is +/- 1, 2, 3, etc. The condition for minima of a single-slit diffraction is the condition for a maxima for a double slit interference when the width of a single slit a is replaced by the separation between the two slits d. This is because in the double slit case, slits are small enough one is considered a single light source and interference of waves from the same slit is ignored, though the minimum condition for single slit is obtained by taking into consideration the interference of waves originating in the same slit.

To calculate the intensity distribution for a pattern produced by a single slit diffraction, find the total field by adding individual point contributions. First, the slit is divided into N small zones of width Δ y = a/N. The convex lens is used to bring parallel light rays to focal point P on the screen. Assuming Δ y is much smaller than lambda, all light from a given zone is in phase. Two adjacent zones have a relative path length (lil)Δ=Δy sin θ. The relative phase shift Δbeta is given by ΔBeta/2ℼ=Δ/lambda=(Δy sin θ)/lambda and Δbeta=2ℼ/lambda * Δy sin θ. If the wavefront from the first point arrives at point P on the screen with electric field E₁=E₁₀sin ω t, the electric field from point 2 adjacent to point 1 will have phase shift Δbeta and the field is E₂=E₁₀sin(ωt+Δbeta). SInce each successive component has the same phase shift relative to the last, the electric field from point N is E_N=E₁₀sin(ωt+(N-1)Δbeta). The total electric field is the sum of individual contributions, E=E₁ to E_N and the total phase shift between points N and 1 is beta=N*Δ*beta=2ℼ/lambda * N Δ y sin θ = 2ℼ/y * a sin θ where N Δ y = a. This can be simplified so the total electric field becomes E=E₁₀[sin(beta/2)/sin(Δbeta/2)]sin(ωt+(N-1)Δbeta/2). I is proportional to the time average of E² and I =I₀/N² * [sin(beta/2)/sin(Δbeta/2)]². The extra factor N² is added to ensure I₀ corresponds to the intensity at central maximum beta=0 (θ=0) and in the limit as Δ beta goes to zero, N sin(Δbeta/2)=NΔbeta/2=beta/2 and intensity becomes I=I₀[sin(beta/2)/(beta/2)]². The condition for minimum intensity is sin θ = m lambda/a.

The intensities for a single slit diffraction are given by I=I₀[sin(ℼ a sin θ/lambda)/(ℼ a sin θ/lambda)]² and intensities for double slit interference are I=I₀cos²(φ/2)=I₀cos²((ℼ d sin θ)/lambda). Given two slits each with width “a” and separated by distance d, the resulting interference pattern for the double slit will include a diffraction pattern due to the individual slit. The intensity of the total pattern is the product of the two functions. I=I₀cos²((ℼ d sin θ)/lambda)[sin(ℼ a sin θ/lambda)/(ℼ a sin θ/lambda)]². The first term is the interference factor which sets the interference substructure and the second is the diffraction factor which acts as an envelope to set limits on the number of interference peaks. The interference maxima occur when d sin θ is m lambda and the first diffraction minimum is a sin θ = lambda. A particular interference maximum with order number m may coincide with the first diffraction minimum. Since, (d sin θ)/(a sin θ)=mlambda/lambda, m=d/a. The mth fringe isn’t seen and the number of fringes on each side of the central is m-1, so the total number in the central diffraction maximum is N=2(m+1)+1=2m-1.

A diffraction grating consists of a large number N of slits each of width “a” and separated from the next by distance d. If the incident light is planar and diffraction spreads light from each slit over a wide angle, so the light from all slits interfere, the relative path difference between each pair of adjacent slits is Δ=d sin θ, similar to the calculation made for the double slit case. If the path difference is equal to an integral of multiple wavelengths, then all slits will constructively interfere and a bright spot will appear on the screen at angle θ. The condition for the principal maxima is d sin θ = m lambda. If the wavelength of the light and location of the m-order maximum are known, the distance d between slits can be deduced. The location of the maxima doesn’t depend on the number of slits, N, but can become sharper and more intense as N increases. The width of the maxima is inversely proportional to N. If an angle θ which initially gives a principal maximum is increased, if there are only two slits, then two waves will still be nearly in phase and produce a broad maxima. With grating and a large number of slits, even though θ may only be slightly deviated from the value producing the maximum, it could be exactly out of phase with light from another slit far away. Grating produces sharper peaks than the two-slit system and gives a more precise wavelength measurement.

Conclusion

While this is the last physics class in my series, the time I’ve spent going through this and classical mechanics and the curiosity I’ve had about quantum mechanics definitely make me want to learn more. It’s been good to get into the habit of reading, but this is definitely a topic where visuals and experiments could help a lot more and there’s still a gap between the materials I read and a lot of this sinking in. Then again, between Classical Mechanics and Electricity and Magnetism, I only spent roughly 4 weeks going through the content for these courses mostly with the intention of learning what content would be covered in a computer science degree, knowing I wouldn’t learn as much as I would in an actual course doing labs. While I’ve felt pretty comfortable comparing my CS knowledge to what I’ve seen so far, physics is definitely an area I knew little about and still have a lot to learn. The last three courses in this series will cover two more computer science courses along with a final course on artificial intelligence. With Software Construction feeling like a pretty complete course on Java and programming in general, I’m curious to see where MIT’s computer science courses go from there.