Questions & explanations
1. Given a system of two coaxial coils with mutual inductance M(x) = M0/(1 + (x/d)^2), where x is separation, find the force between them when currents I1 and I2 are fixed (constant current, not constant flux).
For constant currents, the magnetic energy is W = (1/2) L1 I1^2 + (1/2) L2 I2^2 + M I1 I2. The self terms do not depend on x, so force F = -dW/dx = -I1 I2 dM/dx. Here dM/dx = -2M0 x/(d^2 (1+(x/d)^2)^2). So F = I1 I2 * (2M0 x/(d^2 (1+(x/d)^2)^2)). The force is repulsive if I1 I2 > 0 (currents same direction) and x positive? Actually, sign depends: if x > 0, dM/dx negative, so -I1I2 dM/dx = -I1I2 * negative = positive if I1I2 > 0, meaning force pushes coils apart? Wait, check: For two coils with same direction currents, they attract when brought near. But here if x decreases (closer), M increases, so dM/dx is negative (since M decreases with x). Then -I1I2 dM/dx = -I1I2 * (negative) = positive if I1I2 > 0, meaning force is in positive x direction (increasing separation) — that would be repulsive. That is wrong; same direction currents attract. So careful: For constant current, the force is F = + I1 I2 dM/dx (from Lorentz). Actually, from energy, F = -dW/dx with W including mutual. For constant current, F = - (dM/dx) I1 I2. Yes. So if dM/dx is negative (M decreases with x), then F = - (
2. Find the magnetic field at a point on the axis of a finite wire, a distance d from one end, along the wire's axis.
On the axis of a straight wire, the field is zero because the distance ρ is zero? Actually, the formula B = (μ0 I / (4π ρ)) (sin θ1 + sin θ2) requires ρ not zero. On the axis, ρ = 0, so the formula is not valid directly; the field is zero by symmetry? Wait, for a straight wire, the field is azimuthal; on the axis, the direction is undefined and the magnitude is zero because the contributions from all current elements cancel? Not exactly: Actually, the Biot-Savart law gives zero on the axis of a straight wire because the cross product is zero. So the magnetic field on the axis of a finite straight wire is zero everywhere along the line containing the wire, except perhaps at the wire itself? Typically, the field is zero at points on the wire's axis because the current element and the position vector are parallel. Thus, the answer: The magnetic field is zero at any point lying on the axis of the wire (the line along the wire) because the vector from the current element to the point is parallel to the current, giving zero cross product. This holds for any finite straight wire.
3. Use the Neumann formula to find the mutual inductance between two coaxial circular loops of radius a and b, separated by distance d (with a, b << d).
For two coaxial loops, the distance between elements dl1 and dl2 varies with angle. For large separation d >> a,b, we can approximate r ≈ d. Then the integral ∮∮ dl1·dl2 = (∮ dl1)·(∮ dl2) = (2πa)(2πb) cosθ, where θ is angle between the loops? Actually, coaxial means they are parallel and centered, so dl1 and dl2 are tangential and the dot product is dl1 dl2 cosφ, where φ is angle between the directions. Since loops are parallel and coaxial, the integral over angles gives 0? Wait, careful: For two coaxial circular loops, the mutual inductance is given by M = μ0 π a^2 b^2 / (2 d^3) for a,b << d. Using Neumann: dl1·dl2 = a b dφ1 dφ2 cos(φ1-φ2). The integral over φ1, φ2 gives (2πa)(2πb) * 0? Actually, the cos term averages to zero, but the distance r also depends on φ. The correct result involves elliptic integrals. For simplicity, we can state the approximation: M ≈ μ0 π a^2 b^2 / (2 d^3) when d >> a,b.
4. What is the magnetic field at the center of a uniformly charged sphere rotating about its axis?
For a sphere of radius R, total charge Q, rotating with angular velocity ω, the magnetic field at the center is B = (μ0 Q ω) / (6π R). This is found by treating the rotating charge as a current density J = ρ v, where ρ = Q/(4/3 π R³) and v = ω × r. The Biot-Savart integral gives a uniform field inside the sphere? Actually at the center only. The dipole moment of the sphere is m = (Q ω R²)/5, and B_center = (μ0 m)/(2π R³) = (μ0 Q ω)/(5π R)? Wait, careful: For a uniformly charged rotating sphere, the magnetic field inside is uniform and equals B = (μ0 Q ω)/(6π R) along the axis. Let's verify: Standard result for a sphere of uniform charge density rotating: B_center = (2/3) μ0 σ ω R? Actually better: B = (μ0 Q ω)/(6π R) for a sphere (with volume charge). For a spherical shell, it's different. So use that.
5. How can you find the force between two current-carrying coils using the energy method under constant flux?
Under constant flux, the force is F = -dW/dx, where x is the separation. For two coaxial coils with currents I1, I2, the mutual energy is W = M I1 I2. If currents are kept constant, but flux constant means something else? Actually, constant flux condition means the currents are adjusted to keep flux through each circuit fixed. For two coils, the force is F = I1 I2 dM/dx. If we use constant flux, the energy includes self and mutual, and the force equation becomes F = -(∂W/∂x)_Φ. Usually, for constant current, force is +I1I2 dM/dx. But under constant flux, the sign changes. We need to be consistent. Let's stick to constant flux: F = -∇W, where W includes all magnetic energy. Example: force between two coils is attractive if they increase mutual inductance when brought together.
6. In a material with high permeability, how do B and H compare inside the material?
Inside a high-permeability material like iron, B is much larger than μ0 H. H itself is small because the material's magnetization creates a field that cancels part of the applied H? Actually, H is continuous across boundaries, but B is enhanced. For a given applied current, H is the same as without core, but B is multiplied by the relative permeability. So B can be many times μ0 H. Outside the material, B and μ0 H are almost equal. So the ratio B/H inside is large, but H itself is small? Wait: inside a high μ material, H is actually small if it's a closed loop? Let me rephrase: In a toroidal core, H = NI/l, same as without core, but B = μ0 μr H, so B is large. So indeed B is large, H remains same. The external B field is concentrated in the core.
7. How can you use boundary conditions to explain why a metal shield blocks electric fields inside but not static magnetic fields?
Inside a conductor in static conditions, the electric field is zero. At the boundary, the tangential E must be continuous, so the outside E must also be tangential zero? Actually, the normal D can change due to surface charge. But for static fields, E inside conductor is zero, so the tangential E just outside must also be zero to satisfy continuity. That forces field lines to end on the surface, shielding the interior. For static magnetic fields, the tangential H is continuous if no surface current, but inside a perfect conductor, B cannot change? Actually, static magnetic fields can penetrate conductors, so no shielding. The boundary condition for B normal is continuous, so B can pass through.
8. What does the energy-momentum tensor of the electromagnetic field represent?
The energy-momentum tensor of the electromagnetic field is a mathematical object that describes how energy and momentum are spread out in space and time. It gives the density of energy, the density of momentum, and the flow of energy (like how light carries energy from one place to another). The tensor also tells us about the stress or pressure exerted by the fields. For example, its time-time component is the energy density of the electric and magnetic fields. The off-diagonal parts represent momentum density and energy flow. This tensor is important because its divergence gives the force that the fields exert on charges, ensuring energy and momentum are conserved.
9. Compare the induced EMF method with the transmission line model for calculating dipole impedance.
The induced EMF method gives a more accurate reactance and resistance for thin dipoles, while the transmission line model approximates the dipole as an open-ended transmission line. The transmission line model is simpler: it uses characteristic impedance and length to get impedance, but it ignores radiation losses. The induced EMF method includes radiation resistance explicitly. For a half-wave dipole, the transmission line model gives infinite impedance, which is wrong, whereas the induced EMF method gives 73 ohms. So the induced EMF method is better for predicting input impedance, but the transmission line model is easier for initial estimates.
10. How are electric and magnetic dipole moments similar and different?
Both are vectors that describe the strength and orientation of a dipole source. Both produce fields that decay as 1/r^3 far away, and both experience a torque in an external field (p in E field, m in B field). The main difference is that electric dipoles arise from separated electric charges, while magnetic dipoles arise from circulating currents or intrinsic spin. The units are different: coulomb-meter for electric, ampere-square meter for magnetic. The fields' patterns are similar but with different polarity (electric dipole field lines go from positive to negative; magnetic dipole field lines go from north to south outside the magnet).
11. A flat sheet carries a uniform surface current K in the x-direction. Use the Biot-Savart law to find the direction of the magnetic field just above the sheet.
Consider a small patch on the sheet. The vector K is along x, and r̂ from the patch to a point above the sheet has a vertical component upwards and a horizontal component. The cross product K × r̂ gives a direction out of the page if r̂ is mostly vertical. For an infinite sheet, symmetry gives a field that is uniform and perpendicular to K, parallel to the sheet. Specifically, above the sheet, B points in the y-direction (or -y) depending on orientation. Using the right-hand rule, for K in +x, B above is in +y (if sheet is horizontal). Actually, for a sheet in the xy-plane, K = K x̂, then for a point above (z>0), B is in the y-direction.
12. Compare electromagnetic waves with sound waves.
Electromagnetic waves and sound waves are very different. Sound waves need a medium like air, water, or solid to travel, but electromagnetic waves can travel through empty space. Sound waves are longitudinal, meaning the vibrations are along the direction of travel, while electromagnetic waves are transverse. Sound travels much slower than light: about 340 meters per second in air versus 300,000 kilometers per second for electromagnetic waves. Also, sound waves are mechanical, while electromagnetic waves are caused by electric and magnetic oscillations. You cannot hear electromagnetic waves, but you can see some of them as light.