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 :

Step-by-step solution

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 ) &#8