Question

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?

Short answer

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

Step-by-step solution

Write the given data from the question.

The radius circular ring is $a$.

The mass of circular ring $m$.

The resistance of circular ring is $R$.

Determine the formula to calculate the time taken by the falling the circular ring to attain the terminal velocity.

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 time taken by the falling the circular ring to attain the terminal velocity.

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 $\frac{1}{\alpha }\ln \left(\frac{v_t}{v_t-v}\right)$.