A short bar magnet of magnetic moment 5.25 × 10−2 J T^−1 is placed with its axis

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asked Jan 5, 2018 in Physics by sforrest072 (157,439 points) 61 410 946
edited Mar 4, 2018 by Vikash Kumar

A short bar magnet of magnetic moment 5.25 × 10−2 J T−1 is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45º with earth’s field on
(a) its normal bisector and (b) its axis. Magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.

1 Answer

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answered Jan 5, 2018 by mdsamim (213,225 points) 5 10 15
selected Mar 4, 2018 by Vikash Kumar
 
Best answer

Magnetic moment of the bar magnet, M = 5.25 × 10−2 J T−1
Magnitude of earth’s magnetic field at a place, H = 0.42 G = 0.42 × 10−4 T
(a) The magnetic field at a distance R from the centre of the magnet on the normal bisector is given by the relation:

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