Electromagnetic Induction - Result Question 23
23. A conducting square frame of side ’ $a$ ’ and a long staight 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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(a) $\frac{1}{(2 x-a)^{2}}$
(b) $\frac{1}{(2 x+a)^{2}}$
(c) $\frac{1}{(2 x-a)(2 x+a)}$
(d) $\frac{1}{x^{2}}$
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Answer:
Correct Answer: 23. (c)
Solution:
- (c) Emf induced in side 1 of frame $e_1=B_1 V \ell$
$ B_1=\frac{\mu_o I}{2 \pi(x-a / 2)} $
Emf induced in side 2 of frame $e_2=B_2 V \ell$
$B_2=\frac{\mu_o I}{2 \pi(x+a / 2)}$
Emf induced in square frame $e=B_1 V \ell-B_2 V \ell$
$ =\frac{\mu_0 I}{2 \pi(x-a / 2)} \ell V-\frac{\mu_0 I}{2 \pi(x+a / 2)} \ell V $
or, e $\propto \frac{1}{(2 x-a)(2 x+a)}$
