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The horizontal component of the earth's magnetic field at a certain place is 3.0 xx 10^(-5) T and the direction of the field is form the geographic south to the georgraphic north . A very long straihgt conductor is carrying a steady current of 1 . What is the force par unit length on it when it is placed on horizontal table and the direction of the current is (a) east to west , (b) south to north ? |
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Answer» Solution :`F = 1L xx B` `F= I// B SIN theta` The FORCE per unit length is `f= (F)/(l)` = IB sin `theta` When the current is flowing fromeast to west, `theta = 90^(@)` Hence, `f= IB =1 xx 3 xx 10^(-5) = 3 xx 10^(-5) Nm^(-1)` The direction of the force is downwards. b. When the current is flowing from south to north `theta = 0^(@)` f=0 , Hence there is no force on the conductor. |
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