Solution: Velocity of a Loop upon Switching off Current
1. Magnetic Flux CalculationThe magnetic field at a distance \(x\) from a long straight wire carrying current \(I_{wire}\) is \(B = \frac{\mu_0 I_{wire}}{2\pi x}\). The flux \(\Phi\) through the rectangular loop (from \(x=a\) to \(x=a+b\)) is:
$$ \Phi = \int_{a}^{a+b} \frac{\mu_0 I_{wire}}{2\pi x} (l dx) = \frac{\mu_0 l I_{wire}}{2\pi} \ln\left(\frac{a+b}{a}\right) $$Let the mutual inductance coefficient be \(M = \frac{\mu_0 l}{2\pi} \ln\left(\frac{a+b}{a}\right)\).
2. Induced Current and ForceWhen the current \(I_{wire}\) changes, the induced current in the loop \(i\) is: $$ i = \frac{\varepsilon}{R} = \frac{M}{R} \frac{dI_{wire}}{dt} $$
The net magnetic force on the loop is the difference between forces on the near and far vertical wires: $$ F_{net} = i l [B(a) – B(a+b)] = i l \frac{\mu_0 I_{wire}}{2\pi} \left( \frac{1}{a} – \frac{1}{a+b} \right) $$ $$ F_{net} = i I_{wire} \left[ \frac{\mu_0 l b}{2\pi a(a+b)} \right] $$
3. Impulse and Final VelocitySubstituting \(i = \frac{M}{R} \frac{dI_{wire}}{dt}\) into the force equation: $$ F_{net} = \frac{M}{R} \frac{dI_{wire}}{dt} I_{wire} \left[ \frac{\mu_0 l b}{2\pi a(a+b)} \right] $$ Using the impulse-momentum theorem (\(\int F dt = mv\)): $$ mv = \frac{M}{R} \left[ \frac{\mu_0 l b}{2\pi a(a+b)} \right] \int_{I}^{0} I_{wire} dI_{wire} $$
The integral \(\int I dI = I^2/2\). Substituting \(M\) back: