Two long conductors, separated by a distance `d` carry current `I_(1) and I_(2)` in the same direction . They exert a force `F` on each other. Now the current in one of them is increased to two times and its direction is reversed . The distance is also increased to `3d`. The new value of the force between them is

Options

A. `-F//3`
B. `-2F//3`
C. `F//3`
D. `-2F`

Answer

B

Explanation

In the first case, the force between the two current carrying conductors is
`F_(1) =(mu_(0))/(2pi) (I_(1)I_(2))/(d)l` (It is an attracing force.)
In the second case,
`F_(2) =(mu_(0))/(2pi) [(-(2I_(1))I_(2))/(3d)]l`
It is a repulsive force
`therefore (F_(2))/(F_(1))=-2/3 therefore F_(2)=-2/3 F_(1)or -2/3 F`