Magnetism Question 175
Question: A bar magnet of length 10 cm and having the pole strength equal to 10?3 weber is kept in a magnetic field having magnetic induction equal to $ 4\pi \times {{10}^{-3}} $ Tesla. It makes an angle of 30o with the direction of magnetic induction. The value of the torque acting on the magnet is
[MP PMT 1993]
Options:
A) $ 2\pi \times {{10}^{-7}}N\times m $
B) $ 2\pi \times {{10}^{-5}}N\times m $
C) $ 0.5N\times m $
D) $ 0.5\times 10^{2}N\times m $ ( $ {\mu _{0}}=4\pi \times {{10}^{-7}}weber/amp\times m $ )
Show Answer
Answer:
Correct Answer: A
Solution:
Torque $ \tau =MB _{H}\sin \theta $
$ =0.1\times {{10}^{-3}}\times 4\pi \times {{10}^{-3}}\times \sin 30^{o}={{10}^{-7}}\times 4\pi \times \frac{1}{2} $
$ =2\pi \times {{10}^{-7}}N\times m $
BETA
