# 5.12 (a). A short bar magnet has a magnetic moment of  $0.48 JT^{-1}$. 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,

Given,

The magnetic moment of the bar magnet, m = 0.48 $\dpi{100} JT^{-1}$

Distance from the centre, d = 10 cm = 0.1 m

We know, The magnetic field at distance d, from the centre of the magnet on the axis is:

$B= \frac{\mu_{0} m}{2\pi r^3}$

$\therefore B= \frac{4\pi\times10^{-7}\times 0.48}{2\pi (0.1)^3}$

$\implies$ $B =$ $0.96 \times 10^{-4} T$

Therefore, the magnetic field on the axis, B = 0.96 G

Note: The magnetic field is along the S−N direction (like a dipole!).

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