Setting up equations similar to the last question, we get:
\[\dddot{q}L + \ddot{q} R + \dot{q} \frac{1}{C} + \dot{q} \frac{B^2l^2}{m}=0\]
\[\ddot{i}L + \dot{i} R + i \kappa = 0\]
Here, \(\kappa\) is the same as in last question,
\[\kappa \overset{\text{def}}{=}\frac{1}{C} + \frac{B^2l^2}{m} \]
Solving, and substituting initial conditions, and simplifying, we get:
\[i = \frac{v_0Bl}{\omega L} \sinh (\omega t) e^{-Kt}\tag1\]
Some superconstants used are defined as:
\[\omega \overset{\text{def}}{=} \frac{\sqrt{R^2 - 4 \kappa L}}{2L}\]
\[K \overset{\text{def}}{=} \frac{R}{2L}\]
A point to note is that \(K>\omega\) is always true. This ensures that as \(t\rightarrow \infty\), the current goes to zero. This can be easily proven. Try it!
From eq (1), we can write \(i = \dot{q}\) and integrate it over set limits to get the charge on the capacitor as a function of time:
\[q = \frac{v_0Bl}{L(K^2 - \omega^2)} \left[ 1 - \frac{\sqrt{K^2-\omega^2}}{\omega} \sinh (\alpha + \omega t) e^{-Kt} \right]\tag2\]
Where,
\[\alpha = \sinh^{-1} \left( \frac{\omega}{\sqrt{K^2-\omega^2}}\right)\]
From this, we can find the velocity as a function of time:
\[v = v_0 - \frac{qBl}{m}\tag3\]
Where \(q\) is the function as given in eq (2).
Further, we can find the time at which there exist maximum current in the circuit:
\[t\big|_{\text{max }i} = \frac{1}{\omega} \tanh^{-1} \left( \frac{\omega}{K}\right)\]
Hence, we can find the maximum current in the circuit:
\[i_{\text{max}} = \frac{v_0 Bl}{L} \sqrt{\frac{(K^2-\omega^2)^{ \left( \frac{K^2}{\omega^2} -1 \right)} }{(K^2+\omega^2)^{ \left( \frac{K^2}{\omega^2} +1 \right)}}}\]
As in the previous question, we can also find the total heat generated from the rod due to current flowing through it. Here we make use of the fact that \(\omega < K\) to evaluate an expression in the limit of the integral.
\[\int dH = \frac{v_0^2 B^2l^2 R}{\omega^2 L^2} \int_0^\infty \sinh^2(\omega t) e^{-2Kt} dt\]
\[\implies H = \frac{v_0^2 B^2l^2R}{4L^2K(K^2-\omega^2)}\]