Problem 57

  Two rigid rods are oriented parallel to each other and to the ground. The rods carry the same current in the same direction. The length of each rod is 0.85 m, while the mass of each is 0.073 kg. One rod is held in place above the ground, and the other floats beneath it at a distance of 8.2 x 10^-3 m. Determine the current in the rods.

SOLUTION: The rods are attracted because they carry currents in the same direction. The bottom rod floats because it's in equilibrium - that is, the magnetic and gravitational forces cancel. We know that the magnetic field at a distance, r, produced by a current, I, is

$$B = \frac{\mu_o I}{2 \pi r}$$

and the force exerted on the lower wire by this field is

$$F = ILB\sin\theta$$

$$F = \frac{\mu_o I^2 L \sin\theta}{2 \pi r}$$

where $\theta$ is the angle between the magnetic field at the location of the bottom rod and the direction of the current in the bottom rod. The rods are parallel, and the B-field is perpendicular to the direction of the current, so $\theta$ = 90^o. Setting the gravitational and magnetic forces equal, we get

$$mg = ILB\sin\theta$$

$$mg = \frac{\mu_o I^2 L \sin\theta}{2 \pi r}$$

Solving for the current, we get

$$I = \sqrt{\frac{2 \pi mgr}{\mu_o L}}$$

I = 190 A

which is a pretty hefty current.