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.
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.
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:
$B=\frac{\mu_{0}}{4 \pi} \frac{2 M}{d^{3}}$
Where, $\mu_{0}=$ Permeability of free space $=4 \pi \times 10^{-7} \,T\,m\,A ^{-1}$
$\therefore B=\frac{4 \pi \times 10^{-7} \times 2 \times 0.48}{4 \pi \times(0.1)^{3}}$
$=0.96 \times 10^{-4} T =0.96 \,G$
The magnetic field is along the $S$ - $N$ direction.
$(b)$ The magnetic field at a distance of $10 \,cm$ (i.e., $d=0.1 \,m$ ) on the equatorial of the magnet is given as:
$B=\frac{\mu_{0} \times M}{4 \pi \times d^{3}}$
$=\frac{4 \pi \times 10^{-7} \times 0.48}{4 \pi(0.1)^{3}}$
$=0.48 \,G$
The magnetic field is along the $N-S$ direction.
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