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InterviewSolution
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A short bar magnet has a magnetic moment of `0*48JT^-1`. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of `10cm` from the centre of the magnet on (i) the axis (ii) the equatorial line (normal bisector) of the magnet.A. `0.48xx10^(-4)` T along N-S directionB. `0.28xx10^(-4)` T along S-N directionC. `0.28xx10^(-4)` T along N-S directionD. `0.96xx10^(-4)` T along S-N direction |
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Answer» Correct Answer - D On the axis of the magnet `B=(mu_(0))/(4pi)*(2m)/(d^(3))` Here, `(mu_(0))/(4pi)=10^(-7)"A m"^(-2)` `" "m=0.48" J T"^(-1), d=10` cm = 0.1 m Then, `B=(10^(-7)xx2xx0.48)/((0.1)^(3))` `" "=0.96xx10^(-4)` T along S-N direction. |
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