Magnetism Question 240
Question: A circular coil of 16 turns and radius 10cm carries a current of 0.75 A and rest with its plane normal to an external magnetic field of $ 5.0\times {{10}^{-2}}T $ . The coil is free to rotate about its stable equilibrium position with a frequency of $ 2.0,{{s}^{-1}} $ Compute the moment of inertia of the coil about its axis of rotation.
Options:
A) $ 3.4\times {{10}^{-5}},kg,m^{2} $
B) $ 1.2\times {{10}^{-4}},kg,m^{2} $
C) $ 2.6\times {{10}^{-4}},kg,m^{2} $
D) $ 4.7\times {{10}^{-5}},kg,m^{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] The magnetic moment of the coil is $ M=NIA=16\times 0.75\times \pi \times {{(0.1)}^{2}}=0.377,Am^{2} $ If K be the moment of inertia of the coil about its axis of rotation, then its period of oscillation in a magnetic field B is given by $ T=2\pi \sqrt{\frac{K}{MB}} $ or its frequency v is $ =\frac{1}{T}=\frac{1}{2\pi }\sqrt{\frac{MB}{K}} $ This gives $ K=\frac{MB}{4{{\pi }^{2}}v^{2}} $ Given that $ B=5.0\times {{10}^{-2}}T $ , $ M=0.377,A-m^{2} $ and $ v=2{{s}^{-1}} $
$ \
Therefore ,,,,,,,K=\frac{0.377\times 5.0\times {{10}^{-2}}}{4\times {{(3.14)}^{2}}\times {{(2)}^{2}}}=1.2\times {{10}^{-4}},kgm^{2} $
BETA
