The horizontal component of the earth's magnetic field at a certain place is `3 xx 10^-5 T` and the direction of the field is from the geographic south to the geographic north. A very long straight conductor is carrying a steady current of 1 A. What is the force per unit length on it when it is placed on a horizontal table and the direction of the current is
(a) east to west,
(b) south to north?

Explanation

`vecF=IveclxxvecB`
`F=IlB sin theta`
(a) When the current is flowing from east to west,
` theta=90^@`
Hence `F=IlB=(1A)(1m)(3xx10^-5T)=3xx10^-5N`
The direction of the force is downwards. This direction
may be obtained by either Fleming's left hand rule or the directional property of cross product of vectors.
(b) When the current is flowing from south to north,
` theta=0^@ implies F=0`
Hence, there is no force per unit length on the conductor.