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A magnetic pole of bar magnet with pole strength of 100 Am is 20 cm away from the centre of a bar magnet. Bar magnet has pole-strength of 200 Am and has a length of 5 cm. If the magnetic pole is on the axis of the bar magnet, find the force on the magnetic pole. |
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Answer» Solution :Here, `p= 200` Am `Z= 20 cm = 0.2`m `2l = 5 cm = 0.05 m` `p_1 = 100` Am `F=?` The MAGNETIC moment of bar magnet `m= 2pl = p(2l)` `= 200 xx 0.05 = 10 "Am"^2` The magnetic FIELD at the pole of bar magnet, `B= (mu_0)/( 4pi ) (2m)/( z^3)` `B= (2m)/(z^3) xx 10^(-7)` `= (2 xx 10 xx 10^(-7) )/( (0.2) ^(3) )` `therefore B= 2.5 xx 10^(-4)` T Now the force on the magnetic pole `p_(1) = 100` Am `F= p_(1) B` `= 100 xx 2.5 xx 10^(-4)` `therefore F= 2.5 xx 10^(-2) N` |
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