Magnetism Question 277
Question: The magnetic needle has magnetic moment $ 8.7\times {{10}^{-2}},Am^{2} $ and moment of inertia $ 11.5\times {{10}^{-6}},kgm^{2} $ . It performs 10 complete oscillations in 6.70 s, what is the magnitude of the magnetic field?
Options:
A) 0.012 T
B) 0.120 T
C) 1.200 T
D) 2.10 T
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Magnetic moment, $ M=8.7\times {{10}^{-2}},Am^{2} $ moment of inertia, $ I=11.5\times {{10}^{-6}},kgm^{2} $ Time period of oscillation is $ T=\frac{6.70}{10}=0.6755 $ As, $ T=2\pi \sqrt{\frac{I}{MB}} $
$ B=\frac{4{{\pi }^{2}}I^{2}}{MT^{2}} $
$ \
Therefore ,,,,,,B=\frac{4\times {{(3.14)}^{2}}\times 11.5\times {{10}^{-6}}}{8.7\times {{10}^{-2}}\times {{(0.67)}^{2}}}=0.012,T $
BETA
