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A bar magnet of magnetic moment 0.4Am^(2) is placed in the magnetic meridian with its north pole pointing north. A neutral point is obtained at a distance of 10 cm from the centre of the magnet. If the length of the magnet is also 10 cm, what is the value of horizontal component of earth's field at the place? |
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Answer» Solution :Magnetic moment , `M = 0.4 Am^(2)` Distance of NEUTRAL point from centre of magnet , ` d=10 cm =10 xx 10^(-2)m ` Length of the magnet , ` 2l = 10 cm implies l =5 cm = 5 xx 10^(-2)m` Horizontal component of earth.s FIELD = `B_H` As NORTH pole is pointing north , neutral points are obtained on equatorial line `B_H = B` `B_H = (mu_0)/(4pi)(M)/((d^(2) + l^(2))^(3//2))` `B_H = (4pixx 10^(-7))/(4pi) xx (0.4)/([(10 xx 10^(-2))^(2) + ( 5 xx 10^(-2))^(2)]^(3//2))` ` =2.86 xx 10^(-5)T` |
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