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 :

Earth’s magnetic field at the given place is given by H = 0.36 G

The magnetic field at distance d from the axis of the magnet can be calculated using the following formula –

${{B}_{1}}=\frac{{{\mu }_{0}}2M}{4\pi {{d}^{3}}}=H$

Where \[{{\mu }_{0}}=4\pi \times {{10}^{-7}}Tm{{A}^{-1}}\]

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

${{B}_{2}}=\frac{{{\mu }_{0}}M}{4\pi {{d}^{3}}}=\frac{H}{2}$

Using the relation of B₁, we get –

Total magnetic field is given by – B = B₁ + B₂, therefore –

$B=H+\frac{H}{2}=0.36+0.18$ Hence, 0.54 G is the magnetic field in the direction of earth’s magnetic field.