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 ?

Explanation

`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.