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 [NEET 2015] |
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: C
Solution :
Potential difference across PQ is |
\[{{V}_{P}}-{{V}_{Q}}={{B}_{1}}(a)v=\frac{{{\mu }_{0}}I}{2\pi \left( x-\frac{a}{2} \right)}av\] |
Potential difference across side RS of frame is |
\[{{V}_{S}}-{{V}_{R}}={{B}_{2}}(a)v=\frac{{{\mu }_{0}}I}{2\pi \left( x+\frac{a}{2} \right)}av\] |
Hence, the net potential difference in the loop will be |
\[{{V}_{net}}=({{V}_{P}}-{{V}_{Q}})-({{V}_{S}}-{{V}_{R}})\] |
\[=\frac{{{\mu }_{0}}iav}{2\pi }\left[ \frac{1}{\left( x-\frac{a}{2} \right)}-\frac{1}{\left( x+\frac{a}{2} \right)} \right]\] |
\[=\frac{{{\mu }_{0}}iav}{2\pi }\left( \frac{a}{\left( x-\frac{a}{2} \right)\left( x+\frac{a}{2} \right)} \right)\] |
Thus \[{{V}_{net}}\propto \frac{1}{(2x-a)(2x+a)}\] |
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