A) \[\frac{1}{vR}{{\left[ \frac{{{\mu }_{0}}Iv}{2\pi }In\left( \frac{b}{a} \right) \right]}^{2}}\]
B) \[{{\left[ \frac{{{\mu }_{0}}Iv}{2\pi }In\left( \frac{b}{a} \right) \right]}^{2}}\frac{1}{vR}\]
C) \[{{\left[ \frac{{{\mu }_{0}}Iv}{2\pi }In\left( \frac{b}{a} \right) \right]}^{2}}\frac{v}{R}\]
D) \[\frac{v}{R}{{\left[ \frac{{{\mu }_{0}}Iv}{2\pi }In\left( \frac{b}{a} \right) \right]}^{2}}\]
Correct Answer: A
Solution :
[a] Induced emf \[\int\limits_{a}^{b}{Bvdx}=\int\limits_{a}^{b}{\frac{{{\mu }_{0}}I}{2\pi x}vdx}\] \[\Rightarrow \] Induced emf, \[E=\frac{{{\mu }_{0}}Iv}{2\pi }ln\left( \frac{b}{a} \right)\] \[\Rightarrow \] Power dissipated \[=\frac{{{E}^{2}}}{R}\] Also, power \[=F.v\Rightarrow F=\frac{{{E}^{2}}}{vR}\] \[F={{\left[ \frac{{{\mu }_{0}}Iv}{2\pi }ln\left( \frac{b}{a} \right) \right]}^{2}}\frac{1}{vR}\]You need to login to perform this action.
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