$A$ long straight wire carrying a current of $25 \ A$ rests on a table as shown in the figure. Another wire $PQ$ of length $1 \ m$ and mass $2.5 \ g$ carries the same current but in the opposite direction. The wire $PQ$ is free to slide up and down. To what height $h$ will $PQ$ rise?

(0.51 CM) The magnetic force on the wire $PQ$ due to the current-carrying wire on the table must balance the gravitational force acting on $PQ$ for it to remain in equilibrium at height $h$.
The magnetic field $B$ at a distance $h$ from the long straight wire carrying current $I = 25 \ A$ is given by:
$B = \frac{\mu_0 I}{2 \pi h}$
The magnetic force $F_m$ on the wire $PQ$ of length $l = 1 \ m$ carrying current $I$ is:
$F_m = I l B = I l \left( \frac{\mu_0 I}{2 \pi h} \right) = \frac{\mu_0 I^2 l}{2 \pi h}$
Since the currents are in opposite directions,the magnetic force is repulsive and acts upwards. At equilibrium,this force balances the weight $mg$ of the wire $PQ$:
$mg = \frac{\mu_0 I^2 l}{2 \pi h}$
Rearranging for $h$:
$h = \frac{\mu_0 I^2 l}{2 \pi m g}$
Given: $\mu_0 = 4 \pi \times 10^{-7} \ T \cdot m/A$,$I = 25 \ A$,$l = 1 \ m$,$m = 2.5 \times 10^{-3} \ kg$,$g = 9.8 \ m/s^2$:
$h = \frac{(4 \pi \times 10^{-7}) \times (25)^2 \times 1}{2 \pi \times (2.5 \times 10^{-3}) \times 9.8}$
$h = \frac{2 \times 10^{-7} \times 625}{2.5 \times 10^{-3} \times 9.8} = \frac{1250 \times 10^{-7}}{24.5 \times 10^{-3}} \approx 51.02 \times 10^{-4} \ m$
$h \approx 0.51 \times 10^{-2} \ m = 0.51 \ cm$