InterviewSolution
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InterviewSolution
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A short bar magnet of magnetic moment 5.25 xx 10^(-2)JT^(-1) is placed with its axis perpendicular to the earth's field direction. At what distance from the centre of the magnet on the normal bisector is the resultant field inclined at 45^(@) with the earth's field. Magnitude of earth's field at that place is 0.42 G. [1G=10^(-4)T] |
Answer» Solution :Figure shows the conditions of the problem. SUPPOSE P is the point on the NORMAL bisector of the magnet where the resultant of B (due to magnet) and H is inclined at `45^(@)` with H. This is possible if MAGNITUDES of B and H are the same.  `B = (mu_0)/(4pi) ( M)/(d^(3))`......... for a short magnet ` or d^(3) = (mu_0)/(4pi) (M)/(B) = 10^(-7) xx (5.25 xx 10^(-2))/( 0.42 xx 10^(-4)) = 125 xx 10^(-6)` `:. d = (125 xx 10^(-6))^(1//3) m = 5 xx 10^(-2) m = 5 cm ` |
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