If the distance between two current-carrying parallel wires is halved, then the force between them is

Options

A. Doubled
B. Tripled
C. Quadrupled
D. Halved

Answer

A

Explanation

The force (of attraction or replusion) between two current carrying parallel wires is given by
`F=(mu_(0)I_(1) I_(2)l)/(2pid) i.e.F prop(1)/(d)`
`therefore (F_(2))/(F_(1))=(d_(1))/(d_(2))=(d_(1))/((d_(1))/(2))=2`
`therefore F_(2) =2F_(1)`