Magnetism Question 266
Question: A short bar magnet is placed in the magnetic meridian of the earth with North Pole pointing north. Neutral points are found at a distance of 30 cm from the magnet on the East-West line, drawn through the middle point of the magnet. The magnetic moment of the magnet in $ Am^{2} $ is close to: (Given $ \frac{{\mu _{0}}}{4\pi }={{10}^{-7}} $ in SI units and $ B _{H} $ =Horizontal component of earth’s magnetic field $ =3.6\times {{10}^{-5}},tesla $ )
Options:
A) 14.6
B) 19.4
C) 9.7
D) 4.9
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Here, $ r=30cm=0.3cm $ we know $ \frac{{\mu _{0}}M}{4\pi r^{3}}=B _{H}=3.6\times {{10}^{-5}} $
$ \Rightarrow ,,,M=\frac{3.6\times {{10}^{-5}}}{{{10}^{-7}}}{{(0.3)}^{3}}=9.7,Am^{2} $
BETA
