A bar magnet of length 14 cm is placed in the magnetic meridian with its north pole pointing towards the geographic north pole. A neutral point is obtained at a distance of 18 cm from the centre of the magnet. If BH = 0.4G, the magnetic moment of the magnet is 1G = 10^-4T

I=I0? cos2? $\frac {I_0} {10} = I_0 \cos^2 \theta$10I0? =I0? cos2? $\cos \theta = \frac {1} {\sqrt {10} = 0.31 < \frac {1} {\sqrt {2} \quad \text {which is 0.707}$cos? =10?1? =0.31<2?1? which is 0.707

So,

$\theta > 45^\circ \quad \text {and} \quad 90^\circ - \theta < 45^\circ$? >45? and90? <45?

so only one option is correct i.e. 18.4°

Angle rotated should be

$= 90^\circ - 71.6^\circ = 18.4^\circ$=90?71.6? =18.4?

Answer: (A) 18.4°

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