Magnetism And Matter Ques 22
22. The work done in turning a magnet of magnetic moment $M$ by an angle of $90^{\circ}$ from the meridian, is $n$ times the corresponding work done to turn it through an angle of $60^{\circ}$. The value of $n$ is given by
[1995]
(a) $ 2$
(b) $ 1$
(c) $ 0.5$
(d) $ 0.25$
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Answer:
Correct Answer: 22.(a)
Solution:
- (a) Magnetic moment $=M$; Initial angle through which magnet is turned $(\theta_1)=90^{\circ}$ and final angle through which magnet is turned $(\theta_2)=$ $60^{\circ}$. Work done in turning the magnet through $90^{\circ}(W_1)=MB(\cos 0^{\circ}-\cos 90^{\circ})$ $=M B(1-0)=M B$.
Similarly, $W_2=M B(\cos 0^{\circ}-\cos 60^{\circ})$
$=M B(1-\frac{1}{2})=\frac{M B}{2}$
$\therefore W_1=2 W_2$ or $n=2$.
BETA
