A short bar magnet placed in a horizontal plane has its axis aligned along north-south direction. Null points are found on the axis of the magnet at 20 ~cm from the centre of magnet. The earth's magnetic field at the place is B and angle of dip is 0^ . If the total magnetic field on the normal bisector of the magnet 20 ~cm from the centre of the magnet is 0.6 G , then the magnitude of B is.

Step-by-step solution

At null point (and along the axis), earth's magnetic field and bar's magnetic field are opposite in direction. On equitorial line, bar's mågnetic field is opposite in idirection to its field on the axis. Hence, on equatorial line, the two fields add up As null points are on the axis of the bar magnet, therefore, B₁= ₀ 4 2 M d^3 =H on the equitorial line of magnet at same distance (d), field due to the magnet is B₂= ₀ 4 M d^3 = B₁ 2 = H 2 Therefore, total magnetic field at this point on equitorial line is B=B₂+H= 3