Solution: Loop Entering Magnetic Field (Piecewise Motion)
1. Setup and Basic SHMConsider the loop of width \(b\) and length \(l\). As it enters the field, the side of length \(b\) cuts the magnetic field lines. The flux is \(\Phi = B b x\). The induced current is found via \(L \frac{di}{dt} = \varepsilon = B b v\), which gives \(i = \frac{Bb}{L}x\). The retarding force is \(F = -i b B = -\frac{B^2 b^2}{L}x\).
This force causes Simple Harmonic Motion (SHM) with angular frequency: $$ \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{B^2 b^2 / L}{m}} = \frac{bB}{\sqrt{mL}} $$ The amplitude of this oscillation, if unrestricted, would be \(A = \frac{v_0}{\omega} = \frac{v_0 \sqrt{mL}}{bB}\).
2. Case Analysis based on Initial VelocityThe loop fully enters the field if the maximum displacement \(A\) is greater than the loop’s length \(l\). Threshold velocity condition: \( \frac{v_0 \sqrt{mL}}{bB} > l \implies v_0 > \frac{bBl}{\sqrt{mL}} \).
The loop never fully enters. It performs part of an SHM cycle and returns.
$$ x(t) = \frac{v_0 \sqrt{mL}}{bB} \sin\left( \frac{bBt}{\sqrt{mL}} \right) $$
Phase 1: Entering (\(0 < t < t_0\))
The loop moves with SHM until it is fully inside (displacement \(x=l\)).
$$ x(t) = \frac{v_0 \sqrt{mL}}{bB} \sin\left( \frac{bBt}{\sqrt{mL}} \right) $$
The time \(t_0\) when it fully enters is found by setting \(x=l\):
$$ t_0 = \frac{\sqrt{mL}}{bB} \sin^{-1}\left( \frac{bBl}{v_0 \sqrt{mL}} \right) $$
Phase 2: Fully Inside (\(t > t_0\))
Once fully inside, the flux \(\Phi\) becomes constant (\(Bbl\)). Therefore, the induced EMF becomes zero.
Velocity inside: \(v_{final} = \sqrt{v_0^2 – v_{loss}^2}\).
The position varies linearly with time:
$$ x(t) = \left( \sqrt{v_0^2 – \frac{b^2 B^2 l^2}{mL}} \right) t $$Note: The time variable \(t\) here is total time. The linear equation is valid for \(t > t_0\).