A long horizontal wire `P` carries of `50 A`. It is rigidly fixed. Another fine wire `Q` is placed directly above and parallel to `P`. The weight of wire `Q` is `0.075 N//m` and carries a current of `25 A`. Find the position of wire `Q` from `P` so that the wire `Q` remains suspended due to magnetic repulsion.

Options

A. `10^(-2)m`
B. `(10^(-2))/(2)m`
C. `(10^(-2))/(3)m`
D. `(10^(-2))/(4)m`

Answer

C

Explanation

As force per unit length between two parallel current carrying wires separated by a distance `d` is given by
`(dF)/(dl)=(mu_(0))/(4pi)(2i_(1)i_(2))/(d)`

It is repulsive if the current in the wires are in opposite direction.
In order that wire `Q` remains suspended, the magnetic force must be equal to its weight.
So, `F_(m)=Mg`
`(F)/(L)=(Mg)/(L)implies(mu_(0)i_(1)i_(2))/(2pid)=(Mg)/(L)`
`d=(mu_(0)i_(1)i_(2))/(2pi)=(4pixx10^(-7)xx50xx25)/(2pixx0.075)`
`d=(10^(-2))/(3)m`