Ring and Magnetism (09-08-2020)

The magnetic and gravitational torques about the anchor point should sum to zero. Let the anchor point be at $(x,y,z) = (-R,0,0)$ . The path coordinates and infinitesimal displacement elements are:

$x = R \cos \theta \\ y = R \sin \theta \\ z = 0 \\ dx = - R \sin \theta \, d \theta \\ dy = R \cos \theta \, d \theta \\ dz = 0$

The magnetic force on a piece of the curve is:

$\vec{dF} = I \vec{d \ell} \times \vec{B} \\ \vec{d \ell} = (dx, dy, dz) = (- R \sin \theta \, d \theta, R \cos \theta \, d \theta, 0) \\ \vec{B} = (-B_0, 0, 0)$

Taking the cross product results in the following infinitesimal forces:

$dF_x = 0 \\ dF_y = 0 \\ dF_z = I B_0 \, dy = I B_0 R \cos \theta \, d \theta$

The infinitesimal magnetic torque is:

$\vec{d \tau} = \vec{r} \times \vec{dF} \\ \vec{r} = (R \cos \theta + R, R \sin \theta, 0)$

Taking the cross product results in the following infinitesimal torques:

$d \tau_x = I B_0 R^2 \, \sin \theta \, \cos \theta \, d \theta \\ d \tau_y = -I B_0 R^2 \, \cos^2 \theta \, d \theta -I B_0 R^2 \, \cos \theta \, d \theta \\ d \tau_z = 0$

Integrating these from $\theta = 0$ to $\theta = 2 \pi$ results in:

$\tau_x = 0 \\ \tau_y = - \pi I B_0 R^2 \\ \tau_z = 0$

The gravitational torque is:

$\tau_g = (R,0,0) \times (0,0,-mg) = (0, mgR, 0)$

Observe that these two torques can cancel if the current has the right magnitude.

$\pi I B_0 R^2 = m g R \\ I = \frac{m g}{\pi B_0 R}$