A short bar magnet placed in a horizontal plane has its axis alignedalong the magnetic north-south direction. Null points are found onthe 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 iszero. What is the total magnetic field on the normal bisector of themagnet 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.)

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Solution

Given, the distance of the null point from the magnet is 14 cmand the magnitude of the Earth’s magnetic field is 0.36 G.

The formula to calculate the magnitude of the magnetic field at a distance d on the axis of the magnet is,

B 1 = μ 0 4π ( 2M d 3 ) =H

Here, the permeability of free space is μ 0 , the magnetic moment is M and the distance from the bar magnet is d.

The formula to calculate the magnitude of the magnetic field at a distance d on the equatorial line of a magnet is,

B 2 = μ 0 4π ( M d 3 ) = H 2

The formula to calculate the total magnetic field is,

B= B 1 + B 2 =H+ H 2 = 3 2 H

Substituting the values in the above equation, we get:

B= 3 2 ( 0.36 ) =0.54 G

Thus, the magnitude of the magnetic field is 0.54 G in the direction of the Earth’s magnetic field.

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