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InterviewSolution
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A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth's magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null point (i.e., 14 cm) from the centre of the magnet ? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth's magnetic field.) |
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Answer» Solution :For NULL point on the axis of the magnet at a distance r = 14 CM = 0.14 m from the centre of magnet,the magnetic field of the magnet must NULLIFY the horizontal component of earth.s magnetic field `B_H = 0.36 G = 0.36 XX 10^(-4) T ` (because angle of dip is zero, hence `B_H` = total magnetic field of earth `B_E` ) . `thereforeB_("axial") =(mu_0)/(4pi) * (2m)/(r^3) = B_H = 0.36 xx 10^(-4)` T Now magnetic field at the same distance on the normal bisector of the magnetic will be `B_(eq) =(mu_0)/(4pi) * (m)/(r^3) = (B_("axial"))/(2) = (0.36xx10^(-4))/(2) = 0.18 xx 10^(-4) T` (along the direction of `B_H` ). `therefore ` Total magnetic field on the normal bisector of the magnetic at a distance of 14 cm from the centre of magnet will be `B_("total") = B_(eq) + B_H = 0.18 xx10^(-4)` `=0.54 xx 10^(-4) T = 0.54` G in the direction of earth.s field. |
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