Question

A long, horizontal wire AB rests on the surface of a table and carries a current I. Horizontal wire CD is vertically above wire AB and is free to slide up and down on the two vertical metal guides C and D (Fig. E28.32). Wire CD is connected through the sliding contacts to another wire that also carries a current I, opposite in direction to the current in wire AB. The mass per unit length of the wire CD is l. To what equilibrium height h will the wire CD rise, assuming that the magnetic force on it is due entirely to the current in the wire AB?

Short answer

The magnitude of the height the wire CD reaches is $\frac{{\mu }_0{l}^2}{2\pi g\lambda }$ .

Step-by-step solution

Identification of the concept.

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

 Determination of the height at the wire CD rises

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

$\frac{F}{L}=\frac{{\mu }_0{\parallel }^{\mathrm{\prime }}}{2\pi r}$

Substituting values in the above expression, and we get,

$F_l=\frac{{\mu }_0{I}^{2}L}{2\pi h}$

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

The gravitational force expression is,

$mg=\lambda Lg$

Now, the mechanism of forces acting is,

$\begin{array}{r}{F}_1-mg=0\\ \frac{{\mu }_0{I}^{2}L}{2\pi h}=\lambda Lg\\ h=\frac{{\mu }_0{I}^2}{2\pi g\lambda }\end{array}$

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