Two long conductors `10 cm` apart carry currents in the ration `1:2` in the same direction. The magnetic field midway between them is `2xx10^(-3)` tesla. The force on unit length of any one conductor will be

Options

A. `2pixx10^(-3)N`
B. `2xx10^(-3)N`
C. `piN`
D. `1 N`

Answer

D

Explanation

Magnetic field at their mid-point
`P=(mu_(0))/(2pi((d)/(2)))[2i-i]=(mu_(0)i)/(pid)`
Given `(mu_(0))/(pid)=2xx10^(-3)Tesla`

Substituting `d=0.1 m, i=(10^(3))/(2)A`
Magnetic field at the location of second conductor due to first conductor is `B=(mu_(0)i)/(2pid)`
Force on unit length of second conductor
` = Bxx2i=(mu_(0)i^(2))/(pid)=1newton`