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.
Text Solution
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(i) When the point lies on the axis , then let `B_(1)` be the magnetic field at P , r =10 cm = 0.1 m `therefore " " B_(1) = (mu_(0))/(4pi) . (M)/(r^(3)) = 10^(-7) xx (0.48)/((0.1)^(3))` = `0.48 xx 10^(-4) T = 0.48` G along from N pole to S pole .
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