An arrangment of three parallel staright wires placed perpendcular to plane of paper carrying same current `I` along the same direction is shown in figure. Magnitude of force per unit length on the middle wire `'B'` is given by

Options

A. `(mu_(0)i^(2))/(2 pi d)`
B. `(2 mu_(0) i^(2))/( pi d) `
C. `(sqrt2 mu_(0) i^(2))/( pi d)`
D. `(mu_(0)i^(2))/(sqrt2 pi d) `

Answer

d

Explanation

Let us consider force on l length of wire B. Magnitude of force between BC and AB will be same in magnitude. There will be a force of attraction between pairs of wires because direction of current is same.

`F_(BC) = F_(BA) = (mu_(0)i^(2))/(2 pi d) l`
` F = sqrt 2 F_(BC)`
` = sqrt2 (mu_(0))/(2 pi) i^(2)/d l `
Force per unit length can be written as follows :
` F. = F/l = (mu_(0) i^(2))/(sqrt2 pi d) `
Hence option (d) is correct.