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 gauss 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 the earth’s magnetic field.)

Answer 1

Answer:

Magnetic Field due to magnet is same and contrary to the horizontal component of earth magnetic field at the Null point. Given the data;

d= 14 cm  =0.14 m

H= 0.6 G

B₁= (μ₀*2*m)/(4π*d³)

But the magnetic field is equal to the horizontal component of earth's magnetic field. So the upper modifies;

B₁= (μ₀*2*m)/(4π*d³)

On the equatorial line of magnet at same distance (d) magnetic field due to the magnet.

B₂= (μ₀*m)/(4π*d³) = B₁/2  = H/2

The total magnetic field on equatorial line at this point is

B= B₁+B₂  = H+H/2  = 3/2 H

Since H = 0.36 G

⇒ B= 32*0.36 G  = 0.54 G

Hence Earth's magnetic field is 0.54 Gauss