Question

A short bar magnet with its north pole facing geographic 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 $=B_H$ )

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

Answer 1

Answer: (D)

Initially

Neutral point is obtained on equatorial line and at neutral point $\left|B_H\right|=\left|B_e\right|$
where $B_H=$ Horizontal component of earth's magnetic field, $B_e=$ Magnetic field due to bar magnet on its equatorial line

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