Question

A short bar Magnet with its north pole facing North forms a neutral point P in the horizontal plane. If the magnet is rotated by 90${}^0$ in the horizontal plane, the net magnetic induction at P is:

(Horizontal component of earths magnetic field is $B_H$)

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

Answer 1

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

At P, the equatorial field due to magnet $=B_H$.
After rotation, P lies on the axis of the bar magnet.

Hence, at P, field due to magnet $=2B_H$
Net field is the vector sum of $B_H$ and $2B_H$, both at right angles to each other giving, $B_{net}=\sqrt{5}{B}_H$