Special case, a dipole field. Hard to calculate, but conceptually rich. Suppose

Question

Special case, a dipole field. Hard to calculate, but conceptually rich. Suppose insteed of a uniform field, I have a magnetic dipole (permanent magnet maybe), with magnetic north and the B-field pointing up in the middle of the dipole. For the calculations here, take the area to be0.03 m2 and the total resistance of the loop to be very small, 8 x 10-3 ohms. la. What if I start with the dipole under the table/loop, and let it fall ? At time t-0, the average field in the loop is 120 T (its not uniform though), but three seconds later its far away with essentially no field. Did I induce an electromotive force? What direction was the current ? Use Ohms Law to get the average current. Ib. What if 1 start wth le dipole above the table /loop far away, and let it fall (south-pole first) so that when it is close to the table the average field in the loop is 120 dipole above the table/loop far away, and let it fall Grly how the magretic flux ls changiat, cd dnteretionot the current. s doun Cloccwlse 2. Suppose the magnetic dipole is fixed, underneath the table/loop, and and the dipole is not moving. Instead, you grab the loop/table and yank it upward. Did you induce a current in the loop? If the loop has some resistance and/or a light bulb, you created heat and maybe also light energy. Where did that energy come from ? Reminder of the conceptual hook that connects this last question to all the ones before. You can keep the loop with the current stationary and move the field, or keep the field stationary and move the loop. Faraday's Law is always true, but it apparently comes in a q(v x B) "motional EMF" cause and another "induced EMF cause.

Explanation / Answer

1.

The work shown below the problem is correct for magnitude and direction both.

2.

Yes, current will induce in the loop because there is difference in initial and final magnetic field.

Initial magnetic flux has maximum value because angle between direction of magnetic flux and area of the coil is 0 deg

Calculations are as below,

i=BAcos=BAcos0=BA

For final position =90deg

f=BAcos=BAcos90=0

Induced V=emf = -d/dt = -(0-BA)dt = BA/dt

Thus I=emf/R

Thus current will induce in the loop.

By law of conservation work done to pull the loop upward is transformed into electrical energy or heat energy which flows through the resistance.

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