91影视

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.

$E=Blv$鈥︹��. (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.

$E=IR$ 鈥︹��. (2)

Here, $I$is the current.

The expression to calculate the force is given as follows.

$F=BIl$ 鈥︹��. (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.

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

Let assume$伪=\frac{B^2{l}^2}{mR}$

$\begin{array}{l}\frac{dv}{dt}=g鈭�伪v\\ \frac{dv}{g鈭�伪v}=dt\end{array}$

Integrate both the sides of the above equation.

$鈭�\frac{dv}{g鈭�伪v}=鈭�dt$

Let assume

$\begin{array}{l}g鈭�伪v=u\\ 鈭�伪dv=du\\ dv=鈭�\frac{du}{伪}\end{array}$

Now solve as,

$\begin{array}{l}鈭�鈭�\frac{du}{u伪}=鈭�dt\\ 鈭�伪鈭�dt=鈭�\frac{du}{u}\\ 鈭�伪t=\ln u鈭�\ln A\\ 鈭�伪t=\ln \left(\frac{u}{A}\right)\end{array}$

Solve further as,

$u=A{e}^{鈭�伪t}$

Substitute$g鈭�伪t$for $u$into above equation.

$g鈭�伪v=A{e}^{鈭�伪t}$ 鈥︹��. (4)

At ,$t=0,\text{鈥�}v=0$

$\begin{array}{l}g鈭�伪\left(0\right)=A{e}^{鈭�伪\left(0\right)}\\ g鈭�0=Aa\\ A=g\end{array}$

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

$\begin{array}{l}g鈭�伪v=g{e}^{鈭�伪t}\\ 伪v=g(1鈭�e^{鈭�伪t})\\ v=\frac{g}{伪}(1鈭�e^{鈭�伪t})\end{array}$

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

$\begin{array}{l}v=\frac{g}{\frac{B^2{l}^2}{mR}}(1鈭�e^{鈭�伪t})\\ v=\frac{gmR}{B^2{l}^2}(1鈭�e^{鈭�伪t})\end{array}$ 鈥︹��. (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).

$\begin{array}{l}v=v_t(1鈭�e^{鈭�伪t})\\ 1鈭�e^{鈭�伪t}=\frac{v}{v_t}\\ e^{鈭�伪t}=1鈭�\frac{v}{v_t}\\ e^{鈭�伪t}=\frac{v_t鈭�v}{v_t}\end{array}$

Solve further as,

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

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