Question

A rectangular conducting frame of width $l$ with battery of voltage $V$ is closed with conducting rod of length $l$ and mass $m$. A uniform magnetic field $B$ is present perpendicular to plane. Just after closing the switch, rod starts off with speed $u$. Energy dissipated in this process is

Answer 1

The instantaneous power dissipated is $P_{diss} = \frac{(V - Blu)^2}{R}$. If the question implies total energy dissipated from the start until the rod reaches speed $u$, the answer is $\frac{mVu}{Bl} - \frac{1}{2}mu^2$.

Answer 2

The energy dissipated is the instantaneous power, $P_{diss} = I^2 R$. The current $I$ is given by Ohm's law, $I = \frac{V - \mathcal{E}_{ind}}{R}$, where $\mathcal{E}_{ind} = Blu$ is the induced emf. Thus, $I = \frac{V - Blu}{R}$. Substituting this into the power equation, we get $P_{diss} = \left(\frac{V - Blu}{R}\right)^2 R = \frac{(V - Blu)^2}{R}$. If the question refers to the total energy dissipated until the rod reaches speed $u$, we can use the work-energy theorem. The net force on the rod is $F_{net} = BlI - mg$ (assuming gravity is not acting vertically and the rod is horizontal, or if it is vertical and gravity is balanced by normal force). The initial net force is $F_{net,0} = Bl\frac{V}{R}$ (assuming $u=0$ initially). The final net force is $F_{net,f} = Bl\frac{V-Blu}{R}$. The rate of work done by the battery is $P_{batt} = VI$. The rate of energy dissipation is $P_{diss} = I^2 R$. The rate of change of kinetic energy is $\frac{dK}{dt} = F_{net} u$. By energy conservation, $P_{batt} = P_{diss} + \frac{dK}{dt}$. Integrating over time until speed $u$ is reached: $\int_0^t VI dt = \int_0^t I^2 R dt + \int_0^u m u' du'$. The total energy dissipated is $\int_0^t I^2 R dt$. The work done by the battery is $\int_0^t VI dt$. The change in kinetic energy is $\frac{1}{2}mu^2$. So, $E_{diss} = \int_0^t I^2 R dt = \int_0^t VI dt - \frac{1}{2}mu^2$. The term $\int_0^t VI dt$ is the total energy supplied by the battery. The net emf at any instant is $V_{net} = V - Blu$. The current is $I = \frac{V-Blu}{R}$. The acceleration is $a = \frac{BlI}{m} = \frac{Bl(V-Blu)}{mR}$. Integrating $a$ to find $u(t)$ is complex. A simpler approach is to consider the work-energy theorem in terms of forces and impulses. The net force on the rod is $F_{net} = BlI = \frac{Bl(V-Blu)}{R}$. The equation of motion is $m \frac{du}{dt} = \frac{BlV}{R} - \frac{B^2l^2}{R} u$. This is a first-order linear differential equation. The solution for $u(t)$ is of the form $u(t) = A(1 - e^{-kt})$. The steady-state velocity is $u_{ss} = \frac{V}{Bl}$. The time constant is $\tau = \frac{mR}{B^2l^2}$. So $u(t) = \frac{V}{Bl}(1 - e^{-t/\tau})$. The energy dissipated until speed $u$ is reached is $E_{diss} = \int_0^t I^2 R dt = \int_0^t \left(\frac{V-Blu}{R}\right)^2 R dt = \int_0^t \frac{(V-Blu)^2}{R} dt$. Substitute $u(t)$: $E_{diss} = \int_0^t \frac{(V - Bl\frac{V}{Bl}(1-e^{-t/\tau}))^2}{R} dt = \int_0^t \frac{(V - V(1-e^{-t/\tau}))^2}{R} dt = \int_0^t \frac{(Ve^{-t/\tau})^2}{R} dt = \frac{V^2}{R} \int_0^t e^{-2t/\tau} dt = \frac{V^2}{R} \left[-\frac{\tau}{2} e^{-2t/\tau}\right]_0^t = \frac{V^2\tau}{2R} (1 - e^{-2t/\tau})$. As $t \to \infty$, $u \to V/Bl$, and $E_{diss} \to \frac{V^2\tau}{2R} = \frac{V^2}{2R} \frac{mR}{B^2l^2} = \frac{mV^2}{2B^2l^2}$. This is the total energy dissipated as the rod reaches terminal velocity.

Alternatively, consider the total energy supplied by the battery up to the point where speed $u$ is reached. The total work done by the battery is $W_{batt} = \int_0^t V I dt$. The change in kinetic energy is $\Delta K = \frac{1}{2}mu^2$. The energy dissipated is $E_{diss} = W_{batt} - \Delta K$. The equation of motion is $m \frac{du}{dt} = \frac{BlV}{R} - \frac{B^2l^2}{R} u$. Multiply by $u$: $m u \frac{du}{dt} = \frac{BlV}{R} u - \frac{B^2l^2}{R} u^2$. Integrate from $t=0$ to $t=t_f$ (when speed is $u$): $\int_0^{t_f} m u \frac{du}{dt} dt = \int_0^{t_f} \frac{BlV}{R} u dt - \int_0^{t_f} \frac{B^2l^2}{R} u^2 dt$. $\frac{1}{2} m u^2 = \frac{BlV}{R} \int_0^{t_f} u dt - \frac{B^2l^2}{R} \int_0^{t_f} u^2 dt$. The energy dissipated is $E_{diss} = \int_0^{t_f} I^2 R dt = \int_0^{t_f} \left(\frac{V-Blu}{R}\right)^2 R dt = \int_0^{t_f} \frac{(V-Blu)^2}{R} dt$. Let's use the relation $\int_0^{t_f} V I dt = \int_0^{t_f} (Blu) I dt + \int_0^{t_f} I^2 R dt$. $W_{batt} = E_{induced} + E_{diss}$. $E_{diss} = W_{batt} - E_{induced}$. $W_{batt} = \int_0^{t_f} V \frac{V-Blu}{R} dt = \frac{1}{R} \int_0^{t_f} (V^2 - BluV) dt$. $E_{induced} = \int_0^{t_f} Blu \frac{V-Blu}{R} dt = \frac{1}{R} \int_0^{t_f} (BlVu - B^2l^2u^2) dt$. $E_{diss} = \frac{1}{R} \int_0^{t_f} (V^2 - BluV - BlVu + B^2l^2u^2) dt = \frac{1}{R} \int_0^{t_f} (V^2 - 2BlVu + B^2l^2u^2) dt$. This is not correct.

Let's go back to $W_{batt} = \Delta K + E_{diss}$. $W_{batt} = \int_0^{t_f} VI dt$. $\Delta K = \frac{1}{2}mu^2$. $E_{diss} = \int_0^{t_f} I^2 R dt$. From $m \frac{du}{dt} = \frac{BlV}{R} - \frac{B^2l^2}{R} u$, we can find $u(t)$ and then $I(t)$. The total energy dissipated until the rod reaches speed $u$ is given by $E_{diss} = \frac{mVu}{Bl} - \frac{1}{2}mu^2$. This can be derived by considering the work done by the battery and the change in kinetic energy. The net emf is $V_{net} = V - Blu$. The current is $I = \frac{V-Blu}{R}$. The force is $F = BlI = \frac{Bl(V-Blu)}{R}$. The rate of work done by the battery is $P_{batt} = VI$. The rate of energy dissipation is $P_{diss} = I^2R$. The rate of increase of kinetic energy is $\frac{dK}{dt} = Fu = \frac{Bl(V-Blu)}{R} u$. By energy conservation, $P_{batt} = P_{diss} + \frac{dK}{dt}$. Integrating over time $t_f$ until speed $u$ is reached: $\int_0^{t_f} VI dt = \int_0^{t_f} I^2 R dt + \int_0^{t_f} \frac{Bl(V-Blu)}{R} u dt$. $\int_0^{t_f} VI dt = E_{diss} + \int_0^{t_f} \frac{BlVu - B^2l^2u^2}{R} dt$. The term $\int_0^{t_f} \frac{BlVu}{R} dt$ can be related to the work done by the magnetic force. Consider the work-energy theorem: Work done by battery = Change in KE + Energy dissipated. $W_{battery} = \Delta KE + E_{diss}$. The total work done by the battery is $W_{battery} = \int V I dt$. The induced emf is $\mathcal{E}_{ind} = Blu$. The net emf is $V - Blu$. The current is $I = \frac{V-Blu}{R}$. The work done by the battery is $W_{battery} = \int_0^{t_f} V \frac{V-Blu}{R} dt$. The energy dissipated is $E_{diss} = \int_0^{t_f} I^2 R dt = \int_0^{t_f} \frac{(V-Blu)^2}{R} dt$. The change in kinetic energy is $\Delta KE = \frac{1}{2}mu^2$. We have $m \frac{du}{dt} = \frac{Bl(V-Blu)}{R}$. $m u \frac{du}{dt} = \frac{BlVu - B^2l^2u^2}{R}$. Integrating from $t=0$ to $t_f$: $\int_0^{t_f} m u \frac{du}{dt} dt = \int_0^{t_f} \frac{BlVu - B^2l^2u^2}{R} dt$. $\frac{1}{2} m u^2 = \int_0^{t_f} \frac{BlVu}{R} dt - \int_0^{t_f} \frac{B^2l^2u^2}{R} dt$. Let's consider the work done by the battery: $W_{batt} = \int_0^{t_f} V I dt = \int_0^{t_f} V \frac{V-Blu}{R} dt$. $W_{batt} = \int_0^{t_f} \frac{V^2}{R} dt - \int_0^{t_f} \frac{VBlu}{R} dt$. The term $\int_0^{t_f} \frac{VBlu}{R} dt$ is the work done against the induced emf. We know $E_{diss} = W_{batt} - \Delta KE$. $E_{diss} = \int_0^{t_f} VI dt - \frac{1}{2}mu^2$. Also, $E_{diss} = \int_0^{t_f} I^2 R dt$. Consider the power balance: $VI = I^2R + F u$. $VI = I^2R + \frac{BlI}{R} (BlI/m) = I^2R + \frac{B^2l^2I^2}{mR}$. This is incorrect. $VI = I^2R + \frac{Bl(V-Blu)}{R} u$. $VI = I^2R + \frac{BlVu - B^2l^2u^2}{R}$. Integrate over time: $\int VI dt = \int I^2R dt + \int \frac{BlVu - B^2l^2u^2}{R} dt$. $W_{batt} = E_{diss} + \int \frac{BlVu - B^2l^2u^2}{R} dt$. We have $\frac{1}{2}mu^2 = \int \frac{BlVu}{R} dt - \int \frac{B^2l^2u^2}{R} dt$. Substitute $\int \frac{BlVu}{R} dt = \frac{1}{2}mu^2 + \int \frac{B^2l^2u^2}{R} dt$ into the $W_{batt}$ equation. $W_{batt} = E_{diss} + \left(\frac{1}{2}mu^2 + \int \frac{B^2l^2u^2}{R} dt\right) - \int \frac{B^2l^2u^2}{R} dt$. $W_{batt} = E_{diss} + \frac{1}{2}mu^2$. This is the work-energy theorem. We need to evaluate $W_{batt} = \int_0^{t_f} VI dt$. $W_{batt} = \int_0^{t_f} V \frac{V-Blu}{R} dt = \frac{1}{R} \int_0^{t_f} (V^2 - BluV) dt$. From $m \frac{du}{dt} = \frac{BlV}{R} - \frac{B^2l^2}{R} u$, we have $dt = \frac{mR}{BlV - B^2l^2u} du$. $W_{batt} = \int_0^u V \frac{V-Bl u'}{R} \frac{mR}{BlV - B^2l^2u'} du'$. $W_{batt} = \int_0^u \frac{V(V-Bl u')}{BlV - B^2l^2u'} du'$. $W_{batt} = \int_0^u \frac{V(V-Bl u')}{Bl(V - Blu')} du'$. Let $x = V - Blu'$. $dx = -Bl du'$. $du' = -dx/Bl$. When $u'=0$, $x=V$. When $u'=u$, $x=V-Blu$. $W_{batt} = \int_V^{V-Blu} \frac{V(x)}{Bl(x)} (-\frac{dx}{Bl}) = -\frac{V}{(Bl)^2} \int_V^{V-Blu} dx = -\frac{V}{(Bl)^2} [x]_V^{V-Blu}$. $W_{batt} = -\frac{V}{(Bl)^2} (V-Blu - V) = -\frac{V}{(Bl)^2} (-Blu) = \frac{VBlu}{(Bl)^2} = \frac{Vu}{Bl}$. So, $W_{batt} = \frac{mVu}{Bl}$. Then, $E_{diss} = W_{batt} - \Delta KE = \frac{mVu}{Bl} - \frac{1}{2}mu^2$.