A short bar magnet of magnetic moment 5.25 × 10–2 J T–1 is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45° with earth’s field on (a) its normal bisector and (b) its axis. Magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.
Magnetic moment of the bar magnet, M = 5.25 × 10−2 J T−1
Magnitude of earth’s magnetic field at a place, H = 0.42 G = 0.42 × 10−4
(a) The magnetic field at a distance R from the centre of the magnet on the normal bisector is given by the relation:
When the resultant field is inclined at 45° with earth’s field, B = H (Note: This is the key point of the question, try to sort out this kind of data from the question when you are reading it, so you need not read it again.)
= H = 0.42 × 10−4 T
R3 = 
⇒ 
⇒ R3 = 12.5 x 10-5 m3
⇒ R = 0.05 m = 5 cm
(b) The magnetic field at a distance R1 from the centre of the magnet on its axis is given as:
The resultant field is inclined at 45° with earth’s field.
So, B1 = H
R13 = 
⇒ R13 = 25 x10-5 m3
⇒ R1 = 0.063m = 6.3cm