Q. A short bar magnet with the north pole facing north forms a neutral point a $P$ 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) Zero
B) 2 $B_{H}$
C) $\frac{\sqrt{5}}{2}B_{H}$
D) $\sqrt{5}$ $B_{H}$
Explanation
(d) In Fig. (a), at neutral point $P,$
[image:5_image56.emf]
$B_{H} = \frac{\mu_{0}}{4\pi}\left( \frac{M}{d^{3}} \right)$
In Fig. (b)
Net magnetic induction at $P$ = resultant of $\frac{\mu_{0}}{4\pi}\frac{2\ M}{d^{3}} = 2B_{H}$ along horizontal and $B_{H}$along vertical $= \sqrt{\left( 2B_{H} \right)^{2} + \left( B_{H} \right)^{2}} = \sqrt{5}B_{H}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.