Q. A magnetised wire is bent into an arc of a circle subtending an angle 60° at its centre. Then its magnetic moment is x . If the same wire is bent into an arc of a circle subtending an angle 90° at its centre then its magnetic moment will be
(a) $ \displaystyle \frac {x\sqrt2}{3}$
(b) $\displaystyle \frac {x}{3}$
(c) $ \displaystyle \frac {2\sqrt2 x }{3} $
(d) $ \displaystyle \frac {3 x}{2\sqrt2}$
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Ans: (c)
Sol: When a thin bar magnet of magnetic moment M is bent into an arc of a circle subtending an angle ‘θ’ radians at the centre of the circle, then its new magnetic moment is given by
$ \displaystyle M’ = \frac{2M sin(\theta/2)}{\theta}$ (θ must be in radians)
In first case ,
$ \displaystyle x = \frac{2M sin(\pi/6)}{\pi/3}$
$ \displaystyle x = \frac{3 M}{\pi} $ …(i)
In second case ,
$ \displaystyle x’ = \frac{2M sin(\pi/4)}{\pi/2}$
$ \displaystyle x’ = \frac{2\sqrt{2} M}{\pi} $ …(ii)
From(i)
$ \displaystyle x’ = \frac{2\sqrt{2} x}{3} $