Q83 / 90 · JEE Main 2024 (31 Jan Shift 1) · Physics

A small square loop of wire of side l is placed inside a large square loop of wire of side LL=l2. The loops are coplanar and their centers coincide. The value of the mutual inductance of the system is x×10-7 H, where x=______.

Correct answer: 128

Solution

Figure

Flux linkage for inner loop

ϕ=Bcenter×l2

=4×μ0i4πL2sin45°+sin45°l2

⇒ϕ=22μ0iπLl2

Therefore, mutual inductance will be

M=ϕi=22μ0l2πL=22μ0π

=224ππ×10-7

=82×10-7

=128×10-7 H

Hence, x=128.

Hence, the correct answer is 128.

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