Magnetism Question 265
Question: A 10 cm long bar magnet of magnetic moment $ 1.34,Am^{2} $ is placed in the magnetic meridian with its south pole pointing geographical south. The neutral point is obtained at a distance of 15 cm from the centre of the magnet. Calculate the horizontal component of earth’s magnetic field.
Options:
A) $ 0.12\times {{10}^{-4}},T $
B) $ 0.21\times {{10}^{-4}}T $
C) $ 0.34\times {{10}^{-4}}T $
D) $ 0.87\times {{10}^{-7}}T $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ B=\frac{{\mu _{0}}}{4\pi }\frac{M}{{{(r^{2}+{{\ell }^{2}})}^{3/2}}}=B _{H} $ Given that $ {\mu _{0}}=4\pi \times {{10}^{-7}},T,m{{A}^{-1}} $
$ M=1.34,Am^{2} $
$ r=15,cm=0.15,m $ and $ \ell =5.0,cm=0.05,m $
$ \
Therefore ,,,,,B _{H}={{10}^{-7}}\times \frac{1.34}{{{[{{(0.15)}^{2}}+{{(0.5)}^{2}}]}^{3/2}}} $
$ ={{10}^{-7}}\times \frac{1.34}{0.025\sqrt{0.025}}=0.34\times {{10}^{-4}},T $
BETA
