**Question:**

A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth’s magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null−point (i.e., 14 cm) from the centre of the magnet? (At *null points*, field due to a magnet is equal and opposite to the horizontal component of earth’s magnetic field.)

**Solution:**

Earth’s magnetic field at the given place, *H* = 0.36 G

The magnetic field at a distance *d*, on the axis of the magnet is given as:

$B_{1}=\frac{\mu_{0}}{4 \pi} \frac{2 M}{d^{3}}=H$ …(i)

Where,

$\mu_{0}=$ Permeability of free space

*M* = Magnetic moment

The magnetic field at the same distance *d*, on the equatorial line of the magnet is given as:

$B_{2}=\frac{\mu_{0} M}{4 \pi d^{3}}=\frac{H}{2}$ [Using equation $(i)$ ]

Total magnetic field, $B=B_{1}+B_{2}$

$=H+\frac{H}{2}$

$=0.36+0.18=0.54 \mathrm{G}$

Hence, the magnetic field is 0.54 G in the direction of earth’s magnetic field.