Magnetism Series #3

Consider an inertial frame of reference where the X-axis points horizontally to the right and the Y-axis points upwards. Let the angular velocity of the ring be $\omega$ . The coordinates of the point are, at a general instant of time:

$x = vt + R\sin{\omega t}$ $y = -R - R\cos{\omega t}$

The position vector of this point is:

$\vec{r} = x \ \hat{i} + y \ \hat{j}$ $\implies \vec{r} = (vt + R\sin{\omega t}) \ \hat{i} + (-R - R\cos{\omega t}) \ \hat{j}$

The velocity vector is, therefore:

$\vec{v} = \frac{d\vec{r}}{dt} = (v + R\omega\cos{\omega t}) \ \hat{i} + (R\omega\sin{\omega t}) \ \hat{j}$

Since the ring is purely rolling:

$v = R \omega$

$\implies \vec{v} = (v + v\cos{\omega t}) \ \hat{i} + (v\sin{\omega t}) \ \hat{j}$

The magnetic force on the particle is:

$\vec{F}_B=q(\vec{v} \times \vec{B}) = qvB\left(-\sin{\omega t} \ \hat{i} + (1+\cos{\omega t}) \ \hat{j}\right)$

The gravitational force on the particle is:

$\vec{F}_G = -mg \ \hat{j}$

The total force on the particle is:

$\vec{F} = \vec{F}_B+\vec{F}_G = -qvB\sin{\omega t} \ \hat{i} +\left(qvB (1+\cos{\omega t})-mg\right) \ \hat{j}$

The acceleration of the particle is:

$\vec{a} = \frac{d\vec{v}}{dt} = -R\omega^2\sin{\omega t} \ \hat{i} + R\omega^2\cos{\omega t} \ \hat{j}$

Newton's second law:

$\vec{F} = m\vec{a}$

Equating the X and Y components of LHS and RHS, one gets the following equations:

$qvB = mR\omega^2=mg$ $v = R \omega$

Solving these gives:

$\boxed{R = \frac{v^2}{g}}$ $\boxed{q = \frac{mg}{Bv}}$