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Question
A short bar magnet has a magnetic moment of 0.48 J T−1. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on
- the axis,
- the equatorial lines (normal bisector) of the magnet.
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Solution
Magnetic moment of the bar magnet, M = 0.48 J T−1
- Distance, d = 10 cm = 0.1 m
The magnetic field at distance d, from the centre of the magnet on the axis, is given by the relation:
B = `μ_0/(4pi) (2"M")/"d"^3`
Where,
μ0 = Permeability of free space = 4π × 10−7 T mA−1
∴ B = `(4pi xx 10^-7 xx 2 xx 0.48)/(4pi xx (0.1)^3)`
= 0.96 × 10−4 T
= 0.96 G
The magnetic field is along the S-N direction. - The magnetic field at a distance of 10 cm (i.e., d = 0.1 m) on the equatorial line of the magnet is given as:
B = `(μ_0 xx "M")/(4pi xx "d"^3)`
= `(4pi xx 10^-7 xx 0.48)/(4pi (0.1)^3)`
= 0.48 G
The magnetic field is along the N-S direction.
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