Electro Magnetic Induction And Alternating Currents Question 484
Question: A conducting square frame of side’ a’ and a long staight wire carrying current I are located in the same plane as . The frame moves to the right with a constant velocity ?V?. The emf induced in the frame will be proportional to
Options:
A) $ \frac{1}{{{(2x-a)}^{2}}} $
B) $ \frac{1}{{{(2x+a)}^{2}}} $
C) $ \frac{1}{(2x-a)(2x+a)} $
D) $ \frac{1}{x^{2}} $
Show Answer
Answer:
Correct Answer: C
Solution:
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Emf induced in side 1 of frame $ e_{1}=B_{1}V\ell $
$ B_{1}=\frac{{\mu_{0}}I}{2\pi (x-a/2)} $
Emf induced in side 2 of frame $ e_{2}=B_{2}V\ell $
$ B_{2}=\frac{{\mu_{0}}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}{(2x-a)(2x+a)} $
