A short bar magnet has a magnetic moment of 0.48 JT^-1. Give the direction and magnitude of the magnetic field produced
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 (i) the axis (ii) the equatorial lines (normal bisector) of the magnet.
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Given: m = 0.48 JT-1, r = 10 cm = 0.1 m
To find:
i. Magnetic induction along axis (Ba)
ii. Magnetic induction along equator (Beq)
Formulae:
i. Ba = \(\frac{\mu_0}{4\pi}\,\frac{2m}{r^3}\)
ii. Ba = 2 Beq
Calculation: From formula (i),
Ba = 10-7 × \(\frac{2\times0.48}{10^{-3}}\)
∴ Ba = 0.96 × 10-4 T along S-N direction
From formula (ii),
Beq = \(\frac{0.96\times106-4}{2}\)
∴ Beq = 0.48 × 10-4 T along N-S direction
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