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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 equatorial lines (normal bisector) of the magnet. |
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Answer» Solution :`m = 0.48 JT^(-1),r = 10 cm` a . Field on axial LINE= `(mu_0)/(4pi)(2M)/r^3=(10^(-7)xx2xx0.48)/((10xx10^(-2))^3)=(10^(-7)xx2xx0xx0.48)/(10^(-3))` `=0.96xx10^(-4)T,` in the DIRECTION of `barm` b. Field on equatorial line = `=(mu_0)/(4pi)xxm/r^3=0.48xx10^(-4)T, ` OPPOSITE to `barm` |
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