Electro Magnetic Induction And Alternating Currents Question 511
Question: An infinitesimally small bar magnet of dipole moment $ \vec{M} $ is pointing and moving with the speed v in the $ \hat{x} $ - direction. A small closed circular conducting loop of radius a and negligible self- inductance lies in the y-z plane with its center at x = 0, and its axis coinciding with the x-axis. Find the force opposing the motion of the magnet, if the resistance of the loop is R. Assume that the distance x of the magnet from the center of the loop is much greater than a.
Options:
A) $ \frac{21}{4}\frac{\mu _{0}^{2}M^{2}a^{4}v}{Rx^{8}} $
B) $ \frac{16}{3}\frac{{\mu_{0}}M^{2}a^{2}v^{2}}{Rx^{3}} $
C) $ \frac{3}{23}\frac{{\mu_{0}}Mav^{2}}{Rx^{3}} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
-
$ B=\frac{{\mu_{0}}}{2\pi }\frac{M}{x^{3}} $
$ \therefore $ Flux passing through the coil $ =Ba^{2} $
$ \therefore $ The induced emf in the coil
$ e=\frac{-d\phi }{dt}=-\frac{d}{dt}[ \frac{{\mu_{0}}Ma^{2}}{2}\times \frac{1}{x^{3}} ] $
$ =\frac{3{\mu_{0}}Ma^{2}}{2x^{4}}\frac{dx}{dt}[ \because v=\frac{dx}{dt} ]=\frac{3{\mu_{0}}Ma^{2}}{2x^{4}}\times v $
$ \therefore $ Current in the coil $ =I=\frac{e}{R}=\frac{3{\mu_{0}}Ma^{2}v}{2x^{4}R} $
The magnetic moment of the loop $ M’=I\times A=\frac{3{\mu_{0}}Ma^{2}v}{2x^{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}}Ma^{2}v\times \pi a^{2}}{2x^{4}R}\times \frac{{\mu_{0}}M}{2\pi x^{3}} $
$ \therefore ,U=\frac{3\mu _{0}^{2}M^{2}a^{4}v}{4Rx^{7}} $ Now, $ | {\vec{F}} |=-\frac{dU}{dx} $
$ \therefore F=\frac{21}{4}\frac{\mu _{0}^{2}M^{2}a^{4}v}{Rx^{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.
BETA
