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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 :Here m = 0.48 J `T^(-1)` and r = 10 cm and 0.1 m (a) Magnetic FIELD at a point on its AXIS `B=(mu_0)/(4pi) * (2m)/(r^3) = (10^(-7) XX 2 xx 0.48)/((0.1)^3)` `=9.6 xx 10^(-5)` T along S-N or 0.96 G along S-N direction. (b) Magnetic field at a point at a point on the equatorial line `B =(mu_0)/(4pi)* (m)/(r_3) =(10^(-7) xx 0.48)/((0.1)^3)` `=4.8 xx 10^(-5) ` T along N-S or 0.48 G along N-S direction. |
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