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