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

C

Explanation

Suppose he be the height upto which wire PQ will rise,
Now wire PQ is in equilibrium `F_m =mg`
`=. IIB = mg (because vec(F_m) = vec(II) xx vecB)…….(i)`
Now , `B=(mu_0I)/(2pih)`
From eqns (i) , and (ii),
`N (mu_0I)/(2pi h) = mg " " therefore h =(mu_0 I^2 l)/(2pi mg)`
here, I = 25 A, l=1m, m=2.5 g = `2.5xx10^(-3) kg `
g= 9.8 `ms^(-2) , (mu_0)/(4pi) = 10^(-7) NA^(-2)`
`therefore h(2xx10^(-7) xx (25)xx1)/(2.5xx10^(-3)xx9.8)= 5.1 xx10^(-4) m =0.51 cm`