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 `14cm` from the centre of the magnet. The earth's magnetic field at the plane is `0*36G` 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 points (i.e. 14 cm) from the centre of the magnet? (At null points, field due to a magnet is equal and apposite to the horizontal component of earth's magnetic field).

Explanation

Given, the distance of null points from the centre `d = 14 cm = 14 xx 10^(-2) m`
The horizontal component of the earth.s field `B_H` = 0.36 G
Since the null points are found on the axis of the magnet, so we have:
` (mu_0)/(4pi) (2M)/(d^3) = B_H`
` rArr (mu_0)/(4pi) (M)/(d^3) = (B_H)/(2)` ...(i)
Now on equatorial line at the same distance d from the centre, magnetic field is given by:
`B_1 = (mu_0)/(4pi) (M)/(d^3)`
Using eq (i) `B_1 = (B_H)/(2) = (0.36 xx 10^(-4) )/(2) = 0.18 xx 10^(-4) T`
` therefore ` Total magnetic field at this point on the equatorial line will be
` B = B_1 + B_H`
` = 0.18 xx 10^(-4) + 0.36 xx 10^(-4) `
` = 0.54 xx 10^(-4) T`
` rArr` B = 0.54 G, in the direction of the earth.s field.