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 $14\text{cm}$ from the centre of the magnet. The earth’s magnetic field at the place is $0.36G$ 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., $14\text{cm}$) 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

Given, Earth’s magnetic field, $H=0.36G$

Distance of null point on the axis of magnet, $r=14cm=0.14m$

The magnetic field at distance r from the centre of the magnet on the axis of the magnet, $B_1=\frac{{\mu }_0}{4\pi }\left(\frac{2m}{r^3}\right)$

Here, ${\mu }_0$=permeability of free space $=4\pi \times {10}^{-7}Tm/A$

As magnet is aligned along the magnetic field of earth, at a null point the magnetic field due to the bar magnet is equal and opposite in direction to the earth’s magnetic field.

$\frac{{\mu }_0}{4\pi }\left(\frac{2m}{r^3}\right)=H$

As the magnetic field at distance r, from the centre of the magnet on the equatorial line of the magnet,
$B_2=\frac{{\mu }_0}{4\pi }\left(\frac{m}{r^3}\right)$

$B_2=\frac{H}{2}$
Since the direction of magnetic field due to bar magnet at the equatorial line is opposite to that at the axial line, therefore net field will be the summation of magnetic field due to bar magnet and earth’s magnetic field.
So, the total magnetic field
$B=B_1+B_2$

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

$B=0.36+\frac{0.36}{2}=0.54G$

Final answer: $0.54G$ in the direction of earth’s field