Please provide the detailed solution.
Solution:- 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 T
(a) The magnetic field at a distance R from the centre of the magnet on the normal bisector is given by the relation:
Where,
= Permeability of free space = 4π × 10−7 Tm A−1
When the resultant field is inclined at 45° with earth’s field, B = H
(b) The magnetic field at a distanced
from the centre of the magnet on its axis is given as:
The resultant field is inclined at 45° with earth’s field.
