Electromagnetic field physics question?

Question

A circular coil with a radius 28 cm has 80 turns. An electromagnet produces a uniform magnetic field of strength 1.9 T that is perpendicular to the face of the coil. The current in the electromagnet is turned off, so that the magnetic field decreases to zero in 25 ms.
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What is the average emf induced in the coil during this time? (in volts)

Answer 1

changing B-field produces “induced emf” in coil placed in it

A circular loop of radius (r= 0. 28 m) and n=80 turns when placed in a magnetic field B experiences a magnetic flux.

Flux Phi = B*A cos (p) when p=0 (B and A parallel)
then Phi = B*A

where B (weber = T- m^2), as B is parallel to area vector then p=0. NOW the flux can change either by rotating the coil (dA - through p angle) or by changing field (dB). Here, the latter case is applicable because (p=0) and coil is fixed.

As per Faraday’ law, the changing flux (in given time t) through the coil (of N turns) produces an Induced emf which is equal to

(emf)average = - N [ Delta (phi) ] / Delta (t) ------(1)
(emf)rms = - N [ d (phi) ] / dt ------(2)

The emf given by (1) is average emf (to be found) & negative is just to indicate that it is back emf (which tries to maintain the flux). But when B is varying with time in sine wave then differentials are used, that give “rms” (root mean square) values by which (emf)max can be calculated from (2).

we will use (1)

N = 80, Area = Pi*r^2=3.14*(0.28)^2 = 0.246 m^2,
B1= 1.9 T, B2=0, I think it 25 is milli secs
delta (t) = 25 ms= 0.025 sec

(emf)av
= - N A* [B (2)–B (1)] / Delta (t)
= 80 (0.246* 1.9) / 0.025
= 1495.68 volts