Two long conductors, separated by a distance d carry currents `l_(1)` and `l_(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. `-2F`
B. `F/3`
C. `-(2F)/3`
D. `-F/3`

Answer

C

Explanation

Force acting between two current carrying conductors
`F=(mu_(0))/(2pi)(I_(1)I_(2))/d 1`………..i
Where d= distance between the conductors, l= length of small conductor
Again `F'=(mu_(0))/(2pi)((-2I_(1))(I_(2)))/((3d)) . I=_(mu_(0))/(2pi) (2I_(1)I_(2))/(3d)` .1.............ii
Thus, from eqs i and ii we have `(F')/F=-2/3impliesF'=-2/3F`