Magnetism Question 272
Question: A 30 cm long bar magnet is placed in the magnetic meridian with its north pole pointing south. The neutral point is obtained at a distance of 40cm from the center of the magnet. Find the pole strength of the magnet. The horizontal component of earth’s magnetic field is 0.34 gauss.
Options:
A) 25.7 Am
B) 23.7 Am
C) 28.7 Am
D) 26.7 Am
Show Answer
Answer:
Correct Answer: D
Solution:
[d]
$ \
Therefore ,,,,,,M=\frac{4\pi }{{\mu _{0}}}\frac{B _{H}{{(r^{2}-{{\ell }^{2}})}^{2}}}{2r} $
$ =10^{7}\times \frac{(0.34\times {{10}^{-4}}).{{[{{(0.40)}^{2}}-{{(0.15)}^{2}}]}^{2}}}{2\times 0.40} $
$ =8.0,Am^{2} $ The pole strength of the magnet is, $ m=\frac{M}{2\ell }=\frac{8.0}{0.30}=26.7,A\cdot m $
BETA
