Question

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

• A. $\frac{{\mu }_{0}m}{4\pi d^3}$
• B. $\frac{{\mu }_0\sqrt{2}M}{4\pi d^3}$
• C. $\frac{{\mu }_{0}2\sqrt{2}M}{4\pi d^3}$
• D. $\frac{{\mu }_{0}2M}{4\pi d^3}$

Answer 1

The correct option is B $\frac{{\mu }_0\sqrt{2}M}{4\pi d^3}$

The magnetic field at P due to horizontal magnet above its mid point $B_1=\frac{{\mu }_{0}M}{4\pi d^3}$

The magnetic field at P due to vertical magnet above its mid point $B_2=\frac{{\mu }_{0}M}{4\pi d^3}$

Hence the net magnetic field at point $P=\sqrt{B_1^2+B_2^2}$ $=\frac{{\mu }_0\sqrt{2}M}{4\pi d^3}$