Magnetism Question 85
Question: Moment of inertia of a magnetic needle is 40 gm-cm2 has time period 3 seconds in earth’s horizontal field = $ 3.6\times {{10}^{-5}} $ weber/m2. Its magnetic moment will be
Options:
A) $ 0.5,A\times m^{2} $
B) $ 5,A\times m^{2} $
C) $ 0.250,A\times m^{2} $
D) $ 5\times 10^{2}A\times m^{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ T=2\pi \sqrt{\frac{I}{MB _{H}}} $
$ I=40gm-cm^{2}=400\times {{10}^{-8}}kg-m^{2} $
$ \
Therefore \ 3=2\pi \sqrt{\frac{400\times {{10}^{-8}}}{36\times {{10}^{-6}}\times M}} $
$ \Rightarrow \frac{1}{M}=\frac{9}{4{{\pi }^{2}}}\times \frac{36}{4}\Rightarrow M=0.5\ A\times m^{2} $
BETA
