EMI CYU 8

Part (a): No Cut (Closed Loop)
The frame is perfectly conducting ($R=0$), which implies the magnetic flux through the loop must be conserved.

• Initial state: Vertical. Flux $\Phi = 0$ (Area vector $\perp \vec{B}$).
• If the frame tilts, the flux attempts to change. A current is induced to maintain $\Phi_{net} = 0$.
• This induced current creates a magnetic dipole moment that interacts with $\vec{B}$ to produce a restoring torque opposing the tilt.
• Air resistance dissipates kinetic energy. The frame will settle in the position of minimum potential energy consistent with flux conservation. Since any deviation creates a restoring torque and gravity also acts to pull it vertical, the only equilibrium is the vertical position.

Result (a): Vertical orientation.

Part (b): Cut, Tilted, Closed, then Released
1. Initial Flux Trapping: The frame is tilted to horizontal ($\theta = 90^\circ$). The cut is closed. At this instant, the flux through the frame is maximal: $$ \Phi_{initial} = B \cdot (Area) = B a b $$ Since $R=0$, this flux $Bab$ is conserved throughout the subsequent motion.

2. Equilibrium Position: The frame swings back and settles at a small angle $\theta$ from the vertical. The conservation of flux requires: $$ L I + \Phi_{ext} = \text{Constant} $$ $$ L I + B(ab)\cos(90^\circ – \theta) = B a b $$ $$ L I + B ab \sin\theta = B ab $$ For small angles ($\theta \approx 0$), $\sin\theta \approx 0$. $$ L I \approx B ab \implies I \approx \frac{B ab}{L} $$

3. Torque Balance:
At equilibrium, the magnetic torque balances the gravitational torque.
Gravitational Torque: The center of mass is at distance $a/2$ from the axis (axis is side $b$). $$ \tau_g = mg \left(\frac{a}{2}\right) \sin\theta \approx \frac{mg a \theta}{2} $$ Magnetic Torque: The magnetic moment is $\mu = I A = I ab$. The angle between $\vec{\mu}$ and $\vec{B}$ is $90^\circ – \theta$. $$ \tau_m = \mu B \sin(90^\circ – \theta) = (I ab) B \cos\theta \approx I ab B $$ Substituting $I \approx \frac{Bab}{L}$: $$ \tau_m \approx \left(\frac{Bab}{L}\right) ab B = \frac{B^2 a^2 b^2}{L} $$

4. Solve for $\theta$:
Equating magnitudes of torques: $$ \frac{B^2 a^2 b^2}{L} = \frac{mg a \theta}{2} $$ $$ \theta = \frac{2 B^2 a^2 b^2}{L \cdot mg a} = \frac{2 B^2 a b^2}{mgL} $$