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Magnetism And Matter

Question
CBSEENPH12037724

A short bar magnet has a magnetic moment of 0.48 J T–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.

Solution

Given,
Magnetic moment of the bar magnet, M= 0.48 JT-1
Distance from the centre to the point where magnetic field is produced, r = 10 cm = 0.1 m 

(i) Magnetic field at its axis  Ba = μ04π2Mr3 
Putting values, we get

Ba = 4π×10-7×2×0.484π×(0.1)3
                           Ba = 960 × 10-7 T = 960 × 10-3G 
Ba =0.96 G along the north-south line.  

(ii) Magnetic field along the equatorial line  Be = μ04πMr3

Therefore, putting the values we get, 

 Be = 4π×10-7×0.484π×(0.1)3

 Be = 480 × 10-7 T      = 480 × 10-3G

 Be = 0.48 G along N-S line.