Two short magnets of equal dipole moments M a | vedclass

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(C) The resultant magnetic moment of the two magnets is given by the vector sum of the individual dipole moments. Since they are perpendicular,the net

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$\frac{\mu_0}{4\pi} \frac{2\sqrt{2}M}{d^3}$

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(C) The resultant magnetic moment of the two magnets is given by the vector sum of the individual dipole moments. Since they are perpendicular,the net magnetic moment is $M_{net} = \sqrt{M^2 + M^2} = M\sqrt{2}$.This net magnetic moment acts along the bisector of the angle between the two magnets.The point $P$ lies on the axial line of this equivalent magnetic dipole.The magnitude of the magnetic field on the axial line of a short magnet at a distance $d$ is given by $B = \frac{\mu_0}{4\pi} \frac{2M_{net}}{d^3}$.Substituting $M_{net} = M\sqrt{2}$,we get $B = \frac{\mu_0}{4\pi} \frac{2(M\sqrt{2})}{d^3} = \frac{\mu_0}{4\pi} \frac{2\sqrt{2}M}{d^3}$.

Two short magnets of equal dipole moments $M$ are fastened perpendicularly at their centre (figure). The magnitude of the magnetic field at a distance $d$ from the centre on the bisector of the right angle is

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