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.