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, \[e\propto \frac{1}{(2x-a)(2x+a)}\]You need to login to perform this action.
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