Two short magnets of equal dipole moment M are fastened perpendicularly at their centers. The magnitude of the magnetic field at a distance d from the centre of the bisector of the right angle is

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$\frac{μ_{0}}{4 π} \frac{2 M}{d^{3}}$

$\frac{μ_{0}}{4 π} \frac{2 \sqrt{2} M}{d^{3}}$

$\frac{μ_{0}}{4 π} \frac{\sqrt{2} M}{d^{3}}$

$\frac{μ_{0}}{4 π} \frac{M}{d^{3}}$

answer is D.

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Detailed Solution

The resultant magnetic moment $M^{'} = \sqrt{2} M$

This point will be a point on the axis of the resultant
So the resultant Magnetic field, $B = \frac{μ_{0}}{4 π} \frac{2 M^{'}}{d^{3}} = \frac{μ_{0}}{4 π} \frac{2 \sqrt{2} M}{d^{3}}$

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