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A short bar magnet with magnetic dipole moment 1.6 "Am"^(2) is kept in magnetic meridian in such a way that its north pole is in north direction. In this case, the null (neutral) point is found at a distance of 20 cm from the centre of the magnet. Find the horizontal component of the Earth's magnetic field. |
Answer» Solution :From the figure, it is observed that on the MAGNETIC equator of the magnet, horizontal field lines of the earth.s magnetic field and the magnetic field lines due to the magnet are in mutually opposite directions.  Hence, one finds two points on magnetic equator of the magnet at equal distance from the magnet in such a way, that at these points the above mentioned two magnetic fields are equal in magnitude and opposite in direction. At such points the resultant magnetic field is zero. These points are called neutral or null points. Here, `m= 1.6 "Am"^(2)` Let, the distance of neutral points from the CENTRE of the magnet is `d_1`, `therefore d_(1) = 20 cm = 0.2 m` Now the magnetic field due to a short bar magnet on its equatorial plane `B_1` must equal `B_H` `B_1 = (mu_0)/( 4 pi ) . (m)/( d_(1)^(3) ) = B_(H)` `therefore B_(H) = (10^(-7)xx 1.6) /( (0.2)^(3) ) = 2 xx 10^(-5)` T Now if the bar magnet is kept as in part (b) of the figure, then it is clear that on the magnetic axis, `B_H` and the magnetic field due to the magnet are in mutually opposite directions and so the neutral points are on the axis. Let `d_2` is the distance of such neutral point from the centre of magnet `therefore B_(2) = (mu_0)/( 4PI) . (2m)/( d_(2)^(3) ) = B_H` `therefore d_(2)^(3) = (10^(-7) .2m) /( B_H)= (10^(-7) xx 2 xx 1.6)/( 2 xx 10^(-5) ) ` `= 16 xx 10^(-3) ` `therefore d_(2) = 2.52 xx 10^(-1) m= 25.2` cm |
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