A long straight wire carrying current of `25A` rests on a table as shown in figure. Another wire PQ of length 1m, mass `2*5g` carries the same current but in the opposite direction. The wire PQ is free to slide up and down. To what height will PQ rise?

Options

A. `1.5m`
B. `2.5m`
C. `0.51m`
D. `1m`

Answer

3

Explanation

Suppose `h` be the height upto which wire `PQ` will rise,
Now wise `PQ` is in equilibrium `F_(m)=mg`
`implies IlB=mg ( :' vec(F)_(m)=Ivec(l)xxvec(B)) ..........(i)`
Now,`B=(mu_(0)I)/(2pih)`
From eqns `(i)`, and `(ii)` ,
`Il(mu_(0)I)/(2pih)=mg" " :.h=(mu_(0)I^(2)l)/(2pimg)`
Here, `I=25A,l=1m,m=2.5g=2.5xx10^(-3)kg`
`g=9.8ms^(-2),(mu_(0))/(4pi)=10^(-7)NA^(-2)`
`:. h=(2xx10^(-7)xx(25)xx1)/(2.5xx10^(-3)xx9.8)=51xx10^(-4)m=0.51cm`