A short bar magnet with the north pole facing north forms a neutral point a P in the horizontal plane. If the magnet is rotated by 90° in the horizontal plane, the net magnetic induction at P is (Horizontal component of earth’s magnetic field= $B_{H})$

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$2 B_{H}$

$\frac{\sqrt{5}}{2} B_{H}$

$\sqrt{5} B_{H}$

Zero

answer is D.

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Detailed Solution

at neutral point P,
$B_{H} = \frac{μ_{0}}{4 π} (\frac{M}{d^{3}})$
Net magnetic induction at P = resultant of $\frac{μ_{0}}{4 π} \frac{2 M}{d^{3}} = 2 B_{H}$ along horizontal and

B_H along north
So, the net magnetic induction at P.
$= \sqrt{{(2 B_{H})}^{2} + {(B_{H})}^{2}} = \sqrt{5} B_{H}$

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