A conducting square frame of side a and a long straight wire carrying current I are located in the same plane as shown in the figure. The frame moves to the right with a constant velocity V . The e.m.

Question

A conducting square frame of side a and a long straight wire carrying current I are located in the same plane as shown in the figure. The frame moves to the right with a constant velocity V . The e.m.f induced in the frame (when the centre of the frame is at a distance x from the wire) will be proportional to : NEET 2015 (Phase 1)

TestHub · Accepted Answer

The potential difference across AB is V A - V B = B 1 a . V . ⇒ μ i 2 π x - a 2 a V The potential difference across CD is V C - V D = B 2 a . V B 2 = μ 0 i 2 π x + a 2 V C - V D = μ 0 i 2 π x + a 2 a V Net Potential difference = μ i a V 2 π 1 x - a 2 - 1 x + a 2 ( V A − V B ) − ( V C − V D ) = μ i a 2 π ( 2 a x 2 − a 2 4 ) ∝ 1 4 x 2 - a 2 ∝ 1 2 x + a ( 2 x - a )

Physics - EMI/AC Question with Solution | TestHub

Options:

Doubts & Discussion