InterviewSolution
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InterviewSolution
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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 (a) the axis, (b) the equitorial lines (normal bisector) of the magnet. |
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Answer» Here `M = 0.48 JT^(-1), B = ?` d = 10 cm = 0.1 m (a) On the axis of the magnet `B = (mu_(0))/(4pi).(2M)/(d^(3)) = 10^(-7) xx (2 xx 0.48)/((0.1)^(3))` `= 0.96 xx 10^(T)` along S - N direction (b) On the equitorial line of the magnet `B = (mu_(0))/(4pi) xx (M)/(d^(3)) = 10^(-7) xx (0.48)/((0.1)^(3)) = 0.48 xx 10^(-4)T`, along N - S direction. |
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