Question

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

• A. $0$
• B. $2B_H$
• C. $\frac{\sqrt{5}}{2}{B}_H$
• D. $\sqrt{5}{B}_H$

Answer 1

The correct option is D $\sqrt{5}{B}_H$

Initially

Neutral point obtained on equatorial line and at neutral point
where $B_H$ = Horizontal component of earth’s magnetic field, $B_e$ = Magnetic field due to bar magnet on it’s equatorial line
Finally

Point P comes on axial line of the magnet and at P, net magnetic field $B=\sqrt{B_a^2+B_H^2}$
$=\sqrt{(2B_e{)}^2+(B_H{)}^2}=\sqrt{(2B{H}_H{)}^2+B_H^2}=\sqrt{5}{B}_H$