A) \[\frac{21}{4}\frac{\mu _{0}^{2}{{M}^{2}}{{a}^{4}}v}{R{{x}^{8}}}\]
B) \[\frac{16}{3}\frac{{{\mu }_{0}}{{M}^{2}}{{a}^{2}}{{v}^{2}}}{R{{x}^{3}}}\]
C) \[\frac{3}{23}\frac{{{\mu }_{0}}Ma{{v}^{2}}}{R{{x}^{3}}}\]
D) None of these
Correct Answer: A
Solution :
[a] \[B=\frac{{{\mu }_{0}}}{2\pi }\frac{M}{{{x}^{3}}}\] \[\therefore \] Flux passing through the coil \[=B{{a}^{2}}\] \[\therefore \] The induced emf in the coil
\[e=\frac{-d\phi }{dt}=-\frac{d}{dt}\left[ \frac{{{\mu }_{0}}M{{a}^{2}}}{2}\times \frac{1}{{{x}^{3}}} \right]\] \[=\frac{3{{\mu }_{0}}M{{a}^{2}}}{2{{x}^{4}}}\frac{dx}{dt}\left[ \because \,\,v=\frac{dx}{dt} \right]=\frac{3{{\mu }_{0}}M{{a}^{2}}}{2{{x}^{4}}}\times v\] \[\therefore \] Current in the coil\[=I=\frac{e}{R}=\frac{3{{\mu }_{0}}M{{a}^{2}}v}{2{{x}^{4}}R}\] The magnetic moment of the loop \[M'=I\times A=\frac{3{{\mu }_{0}}M{{a}^{2}}v}{2{{x}^{4}}\,R}\times (\pi {{a}^{2}})\] Now the potential energy of the loop placed in the magnetic field is \[U=-M'B\,\cos {{180}^{o}}=\frac{3{{\mu }_{0}}M{{a}^{2}}v\times \pi {{a}^{2}}}{2{{x}^{4}}R}\times \frac{{{\mu }_{0}}M}{2\pi {{x}^{3}}}\] \[\therefore \,\,\,\,\,U=\frac{3\mu _{0}^{2}{{M}^{2}}{{a}^{4}}v}{4R{{x}^{7}}}\] Now, \[\left| {\vec{F}} \right|=-\frac{dU}{dx}\] \[\therefore \,\,\,\,F=\frac{21}{4}\frac{\mu _{0}^{2}{{M}^{2}}{{a}^{4}}v}{R{{x}^{8}}}\] Since by Newton's third law, action and reaction are equal. Therefore, the above calculated force is acting on the magnet. The direction of the force is in \[-\hat{i}\] direction by Lenz?s law.
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