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.)

Explanation

As null points are on the axis of the magnet, therefore
`B_(1) = (mu_(0))/(4pi)(2M)/(d^(3)) = H`
On the equitorial line of magnet at same distance (d), field due to the magnet is
`B_(2) = (mu_(0))/(4pi)(M)/(d^(3)) = (B_(1))/(2) = (H)/(2)`
`:.` Total Magnetic field at this point on equitorial line is
`B = B_(2) + H = H + (H)/(2) = (3)/(2)H`
`B = (3)/(2) xx 0.36 = 0.54 G`