5. 12 (b). 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 equatorial lines (normal bisector) of the magnet.

Answers (1)

On the equatorial axis,

Distance,d = 10cm = 0.1 m

We know, the magnetic field due to a bar magnet along the equator is:

B= -\frac{\mu_{0} m}{4\pi d^3}

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

\implies B = - 0.48 \times 10^{-4} T

Therefore, the magnetic field on the equatorial axis, B = 0.48 G

The negative sign implies that the magnetic field is along the N−S direction.

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