5.13). 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.)

Answers (1)

Earth’s magnetic field at the given place, B = 0.36 G

The magnetic field at a distance d from the centre of the magnet on its axis is:

B = \mu_{0}m/2\pi d^{3}

And the magnetic field at a distance d' from the centre of the magnet on the normal bisector is:

B' = \mu_{0}m/4\pi d'^3

= B/2  ( since d' = d, i.e same distance of null points.)

Hence the total magnetic field is B + B' = B + B/2 = (0.36 + 0.18) G = 0.54 G

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

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