question_answer

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

A) \[\frac{1}{{{(2x-a)}^{2}}}\]

B) \[\frac{1}{{{(2x+a)}^{2}}}\]

C) \[\frac{1}{(2x-a)(2x+a)}\]

D) \[\frac{1}{{{x}^{2}}}\]

Correct Answer: C

Solution :

[c] 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,  \[e\propto \frac{1}{(2x-a)(2x+a)}\]