Magnetism Question 259
Question: A bar magnet 8 cms long is placed in the magnetic merdian with the N-pole pointing towards geographical north. Two netural points separated by a distance of 6 cms are obtained on the equatorial axis of the magnet. If horizontal component of earth’s field $ =3.2\times {{10}^{-5}}T $ , then pole strength of magnet is
Options:
A) $ 5,ab-amp\times cm $
B) $ 10,ab-amp\times cm $
C) $ 2.5,ab-amp\times cm $
D) $ 20,ab-amp\times cm $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Here, $ 2\ell =8,cm, $
$ \ell =4cm, $
$ d=\frac{6}{2}=3,cm $ . At neutral point, $ H=B=\frac{{\mu _{0}}}{4\pi }\frac{M}{{{(d^{2}+{{\ell }^{2}})}^{3/2}}} $
$ ={{10}^{-7}}\frac{M}{{{(5\times {{10}^{-2}})}^{3}}}=\frac{M}{1250} $
$ \
Therefore ,,,M=1250H=1250\times 3.2\times {{10}^{-5}},Am^{2} $
$ m=\frac{M}{2\ell }=\frac{1250\times 3.2\times {{10}^{-5}}}{8\times {{10}^{-2}}}Am.,,,,=0.5 $
$ Am=0.5\times \frac{1}{10}ab, $
$ amp\times 100,cm $ = 5 ab-amp cm.
BETA
