Refer to Prob. 7.11 (and use the result of Prob. 5.42): How long does is take a falling circular ring (radius $a$, mass $m$, resistance $R$) to cross the bottom of the magnetic field $B$, at its (changing) terminal velocity?

The radius circular ring is $a$.

The mass of circular ring $m$.

The resistance of circular ring is $R$.

The expression to calculate the emf induced in the plate is given as follows.

$E=Blv$……. (1)

Here,$B$ is the magnetic field,$l$ is the length of the segment of the magnetic loop.

The expression to calculate the induced emf in terms of current is given as follows.

$E=IR$ ……. (2)

Here, $I$is the current.

The expression to calculate the force is given as follows.

$F=BIl$ ……. (3)

Calculate the expression for the current

From the equations (1) and (2).

$\begin{array}{c}IR=Blv\\ I=\frac{Blv}{R}\end{array}$

Calculate the upward force acting on the loop.

Substitute $\frac{Blv}{R}$for $I$into equation (3).

$\begin{array}{l}F=B\left(\frac{Blv}{R}\right)l\\ F=\frac{B^2{l}^{2}v}{R}\end{array}$

The upward force, opposed by the gravitational force acting downward.

$\begin{array}{l}{F}_{net}=F_g-F\\ m\frac{dv}{dt}=mg-\frac{B^2{l}^{2}v}{R}\\ \frac{dv}{dt}=g-\frac{B^2{l}^2}{mR}v\end{array}$

Let assume$\alpha =\frac{B^2{l}^2}{mR}$

$\begin{array}{l}\frac{dv}{dt}=g-\alpha v\\ \frac{dv}{g-\alpha v}=dt\end{array}$

Integrate both the sides of the above equation.

$\int \frac{dv}{g-\alpha v}=\int dt$

Let assume

$\begin{array}{l}g-\alpha v=u\\ -\alpha dv=du\\ dv=-\frac{du}{\alpha }\end{array}$

Now solve as,

$\begin{array}{l}-\int \frac{du}{u\alpha }=\int dt\\ -\alpha \int dt=\int \frac{du}{u}\\ -\alpha t=\ln u-\ln A\\ -\alpha t=\ln \left(\frac{u}{A}\right)\end{array}$

Solve further as,

$u=A{e}^{-\alpha t}$

Substitute$g-\alpha t$for $u$into above equation.

$g-\alpha v=A{e}^{-\alpha t}$ ……. (4)

At ,$t=0,v=0$

$\begin{array}{l}g-\alpha \left(0\right)=A{e}^{-\alpha \left(0\right)}\\ g-0=Aa\\ A=g\end{array}$

Substitute $g$for $A$into equation (4).

$\begin{array}{l}g-\alpha v=g{e}^{-\alpha t}\\ \alpha v=g(1-e^{-\alpha t})\\ v=\frac{g}{\alpha }(1-e^{-\alpha t})\end{array}$

Substitute $\frac{B^2{l}^2}{mR}$for$\alpha$into above equation.

$\begin{array}{l}v=\frac{g}{\frac{B^2{l}^2}{mR}}(1-e^{-\alpha t})\\ v=\frac{gmR}{B^2{l}^2}(1-e^{-\alpha t})\end{array}$ ……. (5)

When the loop moves with the internal velocity then the force is balanced by the gravitational force.

$\begin{array}{l}mg=\frac{B^2{l}^2{v}_t}{R}\\ v_t=\frac{mgR}{B^2{l}^2}\end{array}$

Substitute $\frac{mgR}{B^2{l}^2}$for$v_t$ into equation (5).

$\begin{array}{l}v=v_t(1-e^{-\alpha t})\\ 1-e^{-\alpha t}=\frac{v}{v_t}\\ e^{-\alpha t}=1-\frac{v}{v_t}\\ e^{-\alpha t}=\frac{v_t-v}{v_t}\end{array}$

Solve further as,

$\begin{array}{l}{e}^{\alpha t}=\frac{v_t}{v_t-v}\\ \alpha t=\ln \left(\frac{v_t}{v_t-v}\right)\\ t=\frac{1}{\alpha }\ln \left(\frac{v_t}{v_t-v}\right)\end{array}$

Hence,the time taken by the loop to attain the terminal velocity is $\frac{1}{\alpha }\ln \left(\frac{v_t}{v_t-v}\right)$.