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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
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1 (2x - a)2 -
1 (2x + a)2 -
1 (2x - a)(2x + a) -
1 x2
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Correct Option: C
Emf induced in side 1 of frame e1 = B1Vℓ
| B1 = | 2π(x - a/2) |
Emf induced in side 2 of frame e2 = B2Vℓ
| B2 = | 2π(x + a/2) |
Emf induced in square frame
e = B1Vℓ – B2Vℓ
| = | ℓv - | ℓv | ||
| 2π(x - a/2) | 2π(x + a/2) |
| or, e ∝ | (2x - a)(2x + a) |
