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Question

(A) When two short magnets of equal dipole moments $M$ are fastened perpendicularly at their centres,their resultant dipole moment $M^{\prime}$ is giv

Vedclass · Accepted Answer

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

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(A) When two short magnets of equal dipole moments $M$ are fastened perpendicularly at their centres,their resultant dipole moment $M^{\prime}$ is given by the vector sum:$M^{\prime} = \sqrt{M^2 + M^2 + 2 M M \cos 90^{\circ}} = M \sqrt{2}$This resultant dipole moment $M^{\prime}$ acts along the bisector of the right angle.The point at distance $d$ on the bisector lies on the axial line of this resultant dipole $M^{\prime}$.The magnetic field $B$ on the axial position of a short magnetic dipole is given by:$B = \frac{\mu_0}{4 \pi} \frac{2 M^{\prime}}{d^3}$Substituting $M^{\prime} = M \sqrt{2}$ into the formula:$B = \frac{\mu_0}{4 \pi} \frac{2 (M \sqrt{2})}{d^3} = \frac{\mu_0}{4 \pi} \frac{2 \sqrt{2} M}{d^3}$

Two short magnets of equal dipole moments $M$ are fastened perpendicularly at their centres. The magnitude of the magnetic field at a distance $d$ from the centre on the bisector of the right angle is ($\mu_0 =$ Permeability of free space).

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