Chapter 31, Problem 81CP

To determine

The proof that the speed of the bar at time $t$ is given by the formula $v=\frac{\epsilon }{Bd}\left(1-e^{-B^2{d}^{2}t/mR}\right)$.

Answer to Problem 81CP

The proof that the speed of the bar at time $t$ is given by $v=\frac{\epsilon }{Bd}\left(1-e^{-B^2{d}^{2}t/mR}\right)$ is as stated below.

Explanation of Solution

Write the expression to obtain the current produced by the battery.

    $I=\frac{\epsilon }{R}$

Here, $I$ is the current from the battery, $\epsilon$ is the voltage of the battery and $R$ is the resistance.

Write the expression to obtain the applied force on the bar due to the current in the battery.

    $F_{\text{app}}=BId$

Here, $F_{\text{app}}$ is the applied force on the bar due to the current in the battery, $B$ is the magnetic field, $I$ is the current through battery and $d$ is the distance between the two parallel rails.

Substitute $\frac{\epsilon }{R}$ for $I$ in the above equation.

    $F_{\text{app}}=\frac{Bd\epsilon }{R}$                                                                                                                 (I)

The applied force is acting towards the left.

Write the expression to obtain the induced emf.

    $\epsilon \text{'}=-Bdv$

Here, $\epsilon \text{'}$ is the induced emf, $B$ is the magnetic field, $d$ is the distance between the parallel plates and $v$ is the speed of the bar.

Write the expression to obtain the induced current.

    $I\text{'}=\frac{\epsilon \text{'}}{R}$

Here, $I\text{'}$ is the current from the battery, $\epsilon \text{'}$ is the induced emf and $R$ is the resistance.

Substitute $-Bdv$ for $\epsilon \text{'}$ in the above equation.

    $I\text{'}=\frac{-Bdv}{R}$

Write the expression to obtain the induced force due to the induced current on the bar.

    $F_{\text{ind}}=BI\text{'}d$

Here, $F_{\text{app}}$ is the applied force on the bar due to the current in the battery, $B$ is the magnetic field, $I$ is the current through battery and $d$ is the distance between the two parallel rails.

Substitute $\frac{Bdv}{R}$ for $I\text{'}$ in the above equation.

    $\begin{array}{c}{F}_{\text{app}}=Bd\left(\frac{Bdv}{R}\right)\\ =\frac{B^2{d}^{2}v}{R}\end{array}$                                                                                                      (II)

The induced force is acting towards the right.

Write the expression of net force in time $dt$ on the bar.

    $F=m\frac{d{v}_{\text{e}}}{dt}$                                                                                                             (III)

Here, $F$ is the net force in time $dt$ on the bar, $m$ is the mass of the bar, $dv$ is the elemental speed of the bar and $dt$ is the small time interval.

Write the expression to obtain the net force on the bar.

    $F=F_{\text{app}}-F_{\text{ind}}$

Here, $F$ is the net force on the bar, $F_{\text{app}}$ is the applied force on the bar and $F_{\text{ind}}$ is the induced force on the bar.

Substitute $m\frac{d{v}_{\text{e}}}{dt}$ for $F$, $\frac{Bd\epsilon }{R}$ for $F_{\text{app}}$ and $\frac{B^2{d}^{2}v}{R}$ for $F_{\text{ind}}$ in the above equation.

    $\begin{array}{c}m\frac{d{v}_{\text{e}}}{dt}=\frac{Bd\epsilon }{R}-\frac{B^2{d}^{2}v}{R}\\ \frac{d{v}_{\text{e}}}{dt}=\frac{Bd\epsilon }{mR}-\frac{B^2{d}^{2}v}{mR}\\ \frac{d{v}_{\text{e}}}{dt}+\frac{B^2{d}^2}{mR}v=\frac{Bd\epsilon }{mR}\end{array}$                                                                                (IV)

Write the expression of linear differential equation.

    $\frac{d{v}_{\text{e}}}{dt}+Pv=Q$                                                                                                          (V)

Compare equation (IV) and (V).

    $P=\frac{B^2{d}^2}{mR}$

Integrate both sides of the above equation.

    $\begin{array}{l}{\displaystyle \int Pdt}=\frac{B^2{d}^{2}t}{mR}\\ e^{{\displaystyle \int Pdt}}=e^{\frac{B^2{d}^{2}t}{mR}}\end{array}$

Here, $e^{{\displaystyle \int Pdt}}$ is the integrated factor.

Multiply $e^{\frac{B^2{d}^{2}t}{mR}}$ both the side in equation (IV).

    $\begin{array}{l}{e}^{\frac{B^2{d}^{2}t}{mR}}\left(\frac{d{v}_{\text{e}}}{dt}+\frac{B^2{d}^2}{mR}v\right)=e^{\frac{B^2{d}^{2}t}{mR}}\times \frac{Bd\epsilon }{mR}\\ \frac{d}{dt}\left[v_{\text{e}}.e^{\frac{B^2{d}^{2}t}{mR}}\right]=\frac{Bd\epsilon }{mR}{e}^{\frac{B^2{d}^{2}t}{mR}}\end{array}$

Integrate both the sides of the above equation.

    $\begin{array}{l}{e}^{\frac{B^2{d}^{2}t}{mR}}{\displaystyle \int d{v}_{\text{e}}}=\frac{Bd\epsilon }{mR}{\displaystyle \int e^{\frac{B^2{d}^{2}t}{mR}}dt}\\ e^{\frac{B^2{d}^{2}t}{mR}}\left(v\right)=\frac{Bd\epsilon }{mR}\left(\frac{e^{\frac{B^2{d}^{2}t}{mR}}}{\frac{B^2{d}^2}{mR}}\right)+c\end{array}$                                                                                  (VI)

Here, $c$ is the constant term.

Substitute $0$ for $t$ and $v$ in the above equation to find $c$.

    $\begin{array}{c}{e}^{\frac{B^2{d}^{2}t}{mR}}\left(0\right)=\frac{Bd\epsilon }{mR}\left(\frac{e^{\frac{B^2{d}^2\left(0\right)}{mR}}}{\frac{B^2{d}^2}{mR}}\right)+c\\ 0=\frac{Bd\epsilon }{mR}\left(\frac{e^0}{\frac{B^2{d}^2}{mR}}\right)+c\\ c=-\frac{Bd\epsilon }{mR}\left(\frac{1}{\frac{B^2{d}^2}{mR}}\right)\\ =-\frac{Bd\epsilon }{mR}\left(\frac{mR}{B^2{d}^2}\right)\end{array}$

Further solve the above equation.

    $c=-\frac{\epsilon }{Bd}$

Substitute $-\frac{\epsilon }{Bd}$ for $c$ in equation (VI) to find $V$.

    $\begin{array}{l}{e}^{\frac{B^2{d}^{2}t}{mR}}\left(v\right)=\frac{Bd\epsilon }{mR}\left(\frac{e^{\frac{B^2{d}^{2}t}{mR}}}{\frac{B^2{d}^2}{mR}}\right)-\frac{\epsilon }{Bd}\\ e^{\frac{B^2{d}^{2}t}{mR}}\left(v\right)=\frac{\epsilon }{Bd}\left(e^{\frac{B^2{d}^{2}t}{mR}}\right)-\frac{\epsilon }{Bd}\\ e^{\frac{B^2{d}^{2}t}{mR}}\left(v\right)=\frac{\epsilon }{Bd}\left(\left(e^{\frac{B^2{d}^{2}t}{mR}}\right)-1\right)\\ v=\frac{\epsilon }{Bd}\left(1-e^{\frac{-B^2{d}^{2}t}{mR}}\right)\end{array}$

Therefore, the speed of the bar at time $t$ is given by $v=\frac{\epsilon }{Bd}\left(1-e^{-B^2{d}^{2}t/mR}\right)$.