A short bar magnet has a magnetic moment of
Question:

A short bar magnet has a magnetic moment of $0.48 \mathrm{~J} \mathrm{~T}^{-1}$. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of $10 \mathrm{~cm}$ from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.

Solution:

Magnetic moment of the bar magnet, $M=0.48 \mathrm{~J} \mathrm{~T}^{-1}$

(aDistance, 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:

$B=\frac{\mu_{0}}{4 \pi} \frac{2 M}{d^{3}}$

Where,

$\mu_{0}=$ Permeability of free space $=4 \pi \times 10^{-7} \mathrm{Tm} \mathrm{A}^{-1}$

$\therefore B=\frac{4 \pi \times 10^{-7} \times 2 \times 0.48}{4 \pi \times(0.1)^{3}}$

$=0.96 \times 10^{-4} \mathrm{~T}=0.96 \mathrm{G}$

The magnetic field is along the S − N direction.

 

(bThe magnetic field at a distance of 10 cm (i.e., d = 0.1 m) on the equatorial line of the magnet is given as:

$B=\frac{\mu_{0} \times M}{4 \pi \times d^{3}}$

$=\frac{4 \pi \times 10^{-7} \times 0.48}{4 \pi(0.1)^{3}}$

$=0.48 \mathrm{G}$

The magnetic field is along the N − S direction.

 

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