Solution to Question 4
We analyze the forces acting on a bead of mass $m$ and charge $q$ located on the T-frame. Let the bead be at a distance $x$ from the central junction along the cross-bar.
Radius of rotation for a bead at position $x$: $R = \sqrt{l^2 + x^2}$.
Angle $\phi$ with vertical: $\sin \phi = x/R$, $\cos \phi = l/R$.
Forces along the cross-bar (x-direction):
- Centrifugal Force: $F_c = m \omega^2 R$ acting radially outward from the axis of rotation.
Component along the rod: $F_{c,x} = m \omega^2 R \sin \phi = m \omega^2 R (x/R) = m \omega^2 x$. (Outward) - Magnetic Force (Lorentz): Velocity $v = \omega R$ (tangential). $\mathbf{B}$ is perpendicular to the plane.
Force $F_m = q v B = q (\omega R) B$. Direction is radial (towards axis for appropriate signs).
Component along the rod: $F_{m,x} = q \omega R B \sin \phi = q \omega B x$. (Inward) - Coulomb Repulsion: Between beads at $+x$ and $-x$. Distance is $2x$.
$F_e = \frac{1}{4\pi\epsilon_0} \frac{q^2}{(2x)^2}$. (Outward)
Equilibrium Equation: Sum of outward forces = Sum of inward forces.
$$ m \omega^2 x + \frac{q^2}{16 \pi \epsilon_0 x^2} = q \omega B x $$Rearranging:
$$ \frac{q^2}{16 \pi \epsilon_0 x^2} = (q \omega B – m \omega^2) x $$ $$ x^3 = \frac{q^2}{16 \pi \epsilon_0 \omega (q B – m \omega)} $$Taking the cube root:
$$ x = \left[ \frac{q^2}{16 \pi \epsilon_0 \omega^2 ( \frac{qB}{\omega} – m )} \right]^{1/3} $$We need the angular position $\theta$. From the geometry of the T-frame, $\tan \theta = \frac{x}{l}$.
$$ \theta = \tan^{-1} \left( \frac{1}{l} \left[ \frac{q^2}{16 \pi \epsilon_0 \omega^3 ( \frac{qB}{\omega} – m )} \right]^{1/3} \right) $$Simplifying to match the standard form in the key:
$$ \theta = \tan^{-1} \left[ \frac{q^2}{16 \pi \epsilon_0 l^3 (qB\omega – m\omega^2)} \right]^{1/3} $$