EMI BYU 17

Solution: Rod Oscillating with Capacitor and Springs

Let the rod be displaced by \(x\) and moving with velocity \(v\). The motional EMF induced is \(\varepsilon = Blv\). Since this EMF is connected across a capacitor \(C\), the charge on the capacitor is: $$ q = C \varepsilon = C B l v $$ The current in the circuit is the rate of change of charge: $$ i = \frac{dq}{dt} = C B l \frac{dv}{dt} = C B l a $$

The magnetic force on the rod is \(F_m = i l B\). Substituting \(i\): $$ F_m = (C B l a) l B = C B^2 l^2 a $$ The forces acting on the rod are the restoring force from the two springs (\(-2kx\)) and the magnetic force. Note that Lenz’s law implies the magnetic force opposes the change, effectively adding to the inertia. Equation of motion: $$ -2kx – F_m = ma $$ $$ -2kx – C B^2 l^2 a = ma $$ $$ -2kx = (m + C B^2 l^2) a $$

This is the equation of SHM \(a = – \omega^2 x\) where: $$ \omega^2 = \frac{2k}{m + C B^2 l^2} $$ The time period \(T = \frac{2\pi}{\omega}\) is: