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 ( a ) the axis, ( b) the equatorial lines (normal bisector) of the magnet.

Answer 1

Magnetic moment of the bar magnet, M = 0.48 J T^-1 
( a ) 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 =\(\frac{\mu_0}{4\pi}\frac{2M}{d^3}\)
Where,
\(\mu_0\) = Permeability of free space = 4\(\pi\)  \(\times\) 10^-7 ΤmA^-1
∴ B = \(\frac{4\pi\times10^{-7}\times2\times0.48}{4\pi\times(0.1)^3}\)
= 0.96 \(\times\) 10^-4 Τ = 0.96 G
The magnetic field is along the S − N direction.

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( b) 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 = \(\frac{\mu_0\times M}{4\pi\times d^3}\)

= \(\frac{4\pi\times10^{-7}\times \,0.48}{4\pi(0.1)^3}\)

= 0.48 G
The magnetic field is along the N − S direction