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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

1x2

1(2x+a)2

1(2x−a)(2x+a)

1(2x−a)2

Solution:

Induced emf ε=B₁av−B₂av=μ₀I/(2π(x−a/2))av−μ₀I/(2π(x+a/2))av=μ₀Iav/(2π)[1/(x−a/2)−1/(x+a/2)]=μ₀Iav/(2π)[(x+a/2)−(x−a/2)]/[(x−a/2)(x+a/2)]=μ₀Iav/(2π)[a/((x−a/2)(x+a/2))]=μ₀Ia²v/(2π(4x²−a²)) ∝ 1/((2x−a)(2x+a))