91Ó°ÊÓ

The rod will be experiencing a downward pull by gravity. So, the upward force will move the wire CD in an upward direction until it is completely balanced by the gravitational force, which will cause the system to go under equilibrium.

Refer to the image below.

Since the currents are in opposite directions, therefore, the force experienced will be repulsive in nature

Magnetic force per length due to a current-carrying wire is given as,

$\frac{F}{L}=\frac{{\mathrm{μ}}_0{âˆ\yen }^{\mathrm{′}}}{2\mathrm{Ï€}r}$

Substituting values in the above expression, and we get,

$F_l=\frac{{\mathrm{μ}}_0{I}^{2}L}{2\mathrm{π}h}$

Here, L is the length of the wires, and h is the distance between them.

The gravitational force expression is,

$mg=\mathrm{λ}Lg$

Now, the mechanism of forces acting is,

$\begin{array}{r}{F}_1−mg=0\\ \frac{{\mathrm{μ}}_0{I}^{2}L}{2\mathrm{π}h}=\mathrm{λ}Lg\\ h=\frac{{\mathrm{μ}}_0{I}^2}{2\mathrm{π}g\mathrm{λ}}\end{array}$

Thus, h depends on the current I and also $\mathrm{λ}$. The more the current or less $\mathrm{λ}$, the larger the height h is. The value of h is $\frac{{\mathrm{μ}}_0{l}^2}{2\mathrm{Ï€²\mu λ}}$ .