PRS28 Consider the above circular capacitor, and the Amperian loop (radius r) in the plane midway between the plates. When the capacitor is charging, the line integral of the magnetic field around the Amperian loop is 1. Zero: No current crosses the surface spanning the Amperian loop 2. Zero: The magnetic field is perpendicular to the Amperian Loop 3. Non-zero: An electric current flows between the capacitor plates 4. Non-zero: There is time changing electric flux on the surface spanning the Amperian LoopPRS28 GG∫v Bs =µ I +µε dΦ E d⋅ 0 0 0 dtC Answer: 4. When the capacitor is charging up, the line integral of the magnetic field around the Amperian loop is non-zero because there is a time changing electric flux on the flat disc that spans the Amperian LoopPRS28 The plot above shows a side and a top view of a capacitor with charge Q with electric and magnetic fields E and B at time t. The charge Q is: 1. Increasing in time 2. Constant in time. 3. Decreasing in time. 4. Don’t have a clue.PRS28 Answer: 1. The charge Q is increasing in time. The B field is counterclockwise, which means that the current (real & displacement) must be flowing out of the page = up. So there is more charge being carried to the bottom plate.PRS28 The graph shows a plot of the function y = cos( k x ). The value of k is 1. 1/2 2. 1/4 3. π 4. π /2 5. Don’t have a cluePRS28 Answer: 4. k = π /2 λ = 4 Æ k = 2π/λ = π/2 y = cos (π x /2) is 1 at –4, 0, 4, etc.PRS28 The graph shows the E (yellow) and B (blue) fields of a plane wave. This wave is propagating in the 1. +x direction 2. –x direction 3. +z direction 4. –z direction 5. Don’t have a cluePRS28 Answer: 4. –z direction. We can see this because this is the direction of E x B (Yellow x Blue)PRS28 The plot above shows a side and a top view of a capacitor with charge Q with electric and magnetic fields E and B at time t. The charge Q is: 5. Increasing in time 6. Constant in time. 7. Decreasing in time. 8. Don’t have a clue.PRS28 Answer: 1. The charge Q is increasing The direction of the Poynting Flux S (=E x B) inside the capacitor is inward. Therefore electromagnetic energy is flowing inward, and the energy in the electric field inside is increasing. Thus Q must be increasing, since E is proportional to Q.PRS28 The plot above shows a side and a top view of a solenoid carrying current I with electric and magnetic fields E and B at time t. In the solenoid, the current I is: 1. Increasing in time 2. Constant in time. 3. Decreasing in time. 4. Don’t have a clue.PRS28 Answer: 3. The current I is decreasing The Poynting Flux S (= E x B) inside the solenoid is outward from the center of the solenoid. Therefore electromagnetic energy is flowing outward, and the energy in the magnetic field inside is decreasing. Thus I must be decreasing, since B is proportional to
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