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 emf induced in the frame will be proportional to ​

Answer 1

Answer:

The answer will be  (2x-a) -2/(2x+a)

Explanation:

According to figure we can say that as the frame is a square therefore the sides are equal and the side = a

The current located at the similar plane is given as I

and the velocity of the frame is given as v

B1 and B2 are the magnetic fields

let the induced emf of the frame is e

Therefore e = B1(PQ)v -B2(RS)v

                     = μ0I x av/2π(x-a)/2 - μ0I x av/2π(x+a)/2

                     = μ0I av/2π (2/(2x-a) -2/(2x+a))

                    = μ0I av/2π x 2 (2a/(2x-a) -(2x+a))

Therefore we can see that emf e varies with the term of (2x-a) -(2x+a)

Therefore we can say that emf e ∝ 1/ (2x-a) -(2x+a)