Magnetism Question 207
Question: A bar magnet is held perpendicular to a uniform magnetic field. If the couple acting on the magnet is to be halved by rotating it, then the angle by which it is to be rotated is
[CBSE PMT 2000]
Options:
A) 30o
B) 45o
C) 60o
D) 90o
Show Answer
Answer:
Correct Answer: C
Solution:
$ \tau =MB\sin \theta \Rightarrow \tau \propto \sin \theta $
$ \Rightarrow \frac{{\tau _{1}}}{{\tau _{2}}}=\frac{\sin {\theta _{1}}}{\sin {\theta _{2}}}\Rightarrow \frac{\tau }{\tau /2}=\frac{\sin 90}{\sin {\theta _{2}}} $
$ \Rightarrow \sin {\theta _{2}}=\frac{1}{2}\Rightarrow {\theta _{2}}=30^{o} $
Therefore angle of rotation $ =0{\AA}-30=60{}^\circ $
BETA
