The horizontal component of the earth's magnetic field at a certain place is `4.0xx10^(-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 1A. What is the force per unit length on it when it is placed on a horizontal table and the direction of the current is east to west

Explanation

(a)
When I amount of current is passing through a straight conducting wire of length `vecl` (whose direction is taken as the direction of current) placed in a uniform magnetic field `vecB`, magnetic force `vecF` exerted on it is given by Ampere.s law as,
`vecF=I(veclxxvecB)`
`thereforeF=IlBsintheta`
(where `theta` = angle between `veclandvecB`)
Magnetic force per unit length,
`F/l=IBsintheta" "...(1)`
= `IB_(h)sin90^(@)" "(because"Here "B=B_(h)andtheta=90^(@))`
= `(1)(3xx10^(-5))(1)`
= `3xx10^(-5)Nm^(-1)`
(b) Now, in this case as per the statement, conducting wire is placed along `vecB_(h)` and current is passed from south to north and so in equation (1), we should take `theta=0^(@)`. Hence, in this case,
`F/l=0" "(becausetheta=0^(@)rArrsintheta=0)`