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_0i^2)/(2pid)`
B. `(2mu_0i^2)/(pid)`
C. `(sqrt2mu_0i^2)/(pid)`
D. `(mu_0i^2)/(sqrt2pid)`

Answer

D

Explanation

`AB=BC=d`
Magnetic force acting per unit length of wire B due to currents in wires A and B is

`F_1=(mu_0)/(4pi)(2IxxI)/(d)=(mu_0I^2)/(2pid)` along BA
Magnetic force acting per unit length on wire B due to currents in wires B and C is
`F_2=(mu_0I^2)/(2pid)` along BC
`:.` Net magnetic force per unit length on the wire B is
`F=sqrt(F_1^2+F_2^2)=F_1sqrt2 ( :' F_1=F_2)`
`F_2=(mu_0I^2)/(2pid)sqrt2=(mu_0I^2)/(sqrt2pid)`