question_answer

A conducting square frame of side a and a long straight wire carrying current / 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)  \[\frac{1}{{{x}^{2}}}\]                                  

B)  \[\frac{1}{{{(2x-a)}^{2}}}\]

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

D)  \[\frac{1}{(2x-a)(2x+a)}\]

Correct Answer: D

Solution :

 The emf induced in side 1 of frame\[{{e}_{1}}={{B}_{1}}V\ell \] \[{{B}_{1}}=\frac{{{\mu }_{o}}I}{2\pi (x-a/2)}\] The emf induced in side 2 of frame \[{{e}_{2}}={{B}_{2}}V\ell \] \[{{B}_{1}}=\frac{{{\mu }_{o}}I}{2\pi (x+a/2)}\] Net 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,        \[e\propto \frac{1}{(2x-a)(2x+a)}\]