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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 em

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:

Option: 1

\mathrm{\frac{1}{(2 \mathrm{x}-\mathrm{a})^2}}


Option: 2

\mathrm{\frac{1}{(2 \mathrm{x}+\mathrm{a})^2}}


Option: 3

\mathrm{\frac{1}{(2 x-a)(2 x+a)}}


Option: 4

\mathrm{\frac{1}{\mathrm{x}^2}}


Answers (1)

best_answer

\mathrm{As\: the\: emf\: induced =\mathrm{B}_1 \mathrm{vl}-\mathrm{B}_2 \mathrm{vl}}

\mathrm{=\frac{\mu_0 I}{2 \pi(x-a / 2)} l v-\frac{\mu_0 I}{2 \pi(x+a / 2)} l v \propto \frac{1}{(2 x-a)(2 x+a)} }

 

 

Posted by

Suraj Bhandari

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