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.)

Answer 1

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:
$B_1$ = $\frac{\mu_0}{4\pi}\frac{2M}{d^3}$= 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:
$B_2$ = $\frac{\mu_0M}{4\pi d^3}$ = $\frac{H}{2}$ [Using equation (i)]
Total magnetic field, $B$ = $B_1+B_2$
=$H+\frac{H}{2}$
= 0.36+0.18=0.54 G

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