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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 velo

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 x+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

\text { Emf induced in side } 1 \text { of frame } e_1=B_1 V l

\mathrm{B_1=\frac{\mu_0 I}{2 \pi(x-a / 2)}}

\mathrm{\text { Emf induced in side } 2 \text { of frame } e_2=B_2 \mathrm{Vl}}

\mathrm{B_2=\frac{\mu_0 I}{2 \pi(x+a / 2)}}

\mathrm{Emf\: induced\: in\: square \: frame\: e=B_p \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 }

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

Posted by

Suraj Bhandari

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