A long horizontally fiexed wire carries a current of 100 A directely above and parallel to it is a fine wire that carries a current of 20 A and weight 0.04 N/m The distance between the two wires for which the upper wire is just supported by magnetic repulsion is

Options

A. `10^(-2)mm`
B. `10^(-2)cm`
C. `10^(-2)m`
D. `10^(-2) km`

Answer

C

Explanation

As we know that force / length between a two current carrying wire is given by
`(dF)/(dI)=(mu_(0))/(2pi)(i_(1)i_(2))/(d)`
so accroding to question `(dF)/(di)=0.04`
so `0.04 =(mu_(0))/(2pi)(100xx20)/(d)`
`rarr d=(mu_(0))/(2pi)(100xx20)/(0.04)`
`rarr as (mu_(0))/(2pi)=2xx10^(-7)`
So `d=(2xx10^(-7)xx100xx20)/(0.04)`
so `d=(2xx10^(-7)xx100xx20)/(0.04)`
`10^(-2) m`