Solution to Question 3
Physics Principle: The particle is in a non-inertial rotating frame. The forces acting on the particle of mass $m$ and charge $q$ are:
- Tension $T$ (Inward towards center)
- Centrifugal Force $F_c = m\omega^2 R_{path}$ (Outward)
- Magnetic Force (Lorentz) $F_m = qvB$. Since $v = \omega R_{path}$ (tangential), the magnetic force is radial. Direction depends on cross product $\mathbf{v} \times \mathbf{B}$.
The radius of the path is $R_{path} = l + r$ (distance from center $C$ to the nail $O$ plus thread length, assuming the thread aligns radially). Wait, the nail is at distance $r$ from center. Thread length $l$. The particle $P$ is at distance $R = r + l$ from the center.
Force Equation
Assuming the magnetic field is directed such that the magnetic force is inward (for $q>0, \omega>0, \mathbf{B}\uparrow$): $F_m = q(\omega R)B$ (Inward).
The equation of motion for equilibrium in the rotating frame is:
$$ T_{in} + F_{m, in} = F_{c, out} $$ $$ T + q\omega R B = m \omega^2 R $$ $$ T = m \omega^2 R – q \omega R B $$However, Tension cannot be negative. If $m \omega^2 R < q \omega R B$, the string goes slack, so $T=0$.
Critical angular velocity where $T=0$:
$$ m \omega^2 R = q \omega R B \implies \omega_0 = \frac{qB}{m} $$For $0 \le \omega \le \frac{qB}{m}$, the Net Force is Inward. The particle moves to position $R = r – l$ (closest approach to center). Tension balances the net inward force.
$$ T = F_{mag} – F_{cent} = q\omega B(r-l) – m\omega^2(r-l) $$For $\omega > \frac{qB}{m}$, the Centrifugal Force dominates. The particle moves to $R = r + l$. Tension balances the net outward force.
$$ T = F_{cent} – F_{mag} = m\omega^2(r+l) – q\omega B(r+l) $$