Page No 202: - Chapter 5 - Magnetism & Matter - Exercise Solutions - Ncert Solutions class 12 - Physics
Updated On 26-08-2025 By Lithanya
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Question 5.15:
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 45º 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.
Answer:
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.
