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 $\text{P}$ is- [Take $B_H$ as the horizontal component of earth's magnetic field]

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

Answer 1

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

At the neutral point, horizontal component of earth's field $B_H$ exactly balances field due to the magnet.

$B_H=B_{eq}$



When rotated ${90}^o$ in the horizontal plane




$B_R=\sqrt{B_H^2+B_{ax}^2}[\because B_{ax}=2B_{eq}]$


$B_R=\sqrt{B_H^2+(2B_{eq}{)}^2}[\because B_{eq}=B_H]$

$B_R=\sqrt{B_H^2+4B_H^2}$

$B_R=\sqrt{5}{B}_H$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.