Question

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

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

Answer 1

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

When the north pole of short bar magnet is facing North pole of the earth, at the neutral point P, which is on equatorial line.
$B_H=\frac{{\mu }_{0}M}{4\pi d^3}=B_1$ ............(1)
When the magnet is rotated by ${90}^o$, the magnetic induction at P which is on axial line,
$B_H=\frac{{\mu }_{0}2M}{4\pi d^3}=B_2$ ............(2)
Therefore, net magnetic induction at P is
$B_{net}=\sqrt{(B_1^2+B_2^2)}$
$B_{net}=\sqrt{(1^2+2^2)}{B}_H=\sqrt{5}{B}_H$