Question

A rod of mass $m$ and resistance $r$ is placed on fixed, smooth and resistanceless conducting rails which is closed by a resistance $R$ and it is projected with an initial velocity $u$. Find its velocity at any instant as a function of time $\left(t\right)$.

• A. $u{e}^{-B^2{l}^{2}t/(R+r)m}$
• B. $u{e}^{-B^2{l}^{2}t/2m(R+r)}$
• C. $u{e}^{-2B^2{l}^{2}t/(R+r)m}$
• D. $\frac{u}{2}{e}^{-2B^2{l}^{2}t/(R+r)m}$

Answer 1

The correct option is A $u{e}^{-B^2{l}^{2}t/(R+r)m}$

Let at any instant the velocity of the rod is $v$.

The emf induced in the rod will be, $E=Bvl$

The equivalent circuit is shown in the figure.


$\therefore$ Current in the circuit is,

$i=\frac{Bvl}{R+r}$

From Lenz's law, the direction of current in the circuit will be clockwise.

Due to this current, a magnetic force will act on the rod, opposite to the motion of the rod.

The force is given by, $F=ilB$

Using,
$F=ma=\frac{mdv}{dt}$

$\Rightarrow m\frac{dv}{dt}=ilB=\frac{B^2{l}^{2}v}{R+r}[\because i=\frac{Blv}{R+r}]$

Since, acceleration is acting in the opposite direction of the velocity,

$\therefore \frac{dv}{dt}=-\frac{B^2{l}^{2}v}{m(R+r)}$

Integrating the above equation within proper limit,

${\int }_u^v\frac{dv}{v}={\int }_0^t\frac{-B^2{l}^2}{(R+r)m}dt$

$\Rightarrow \ln \left(\frac{v}{u}\right)=\frac{-B^2{l}^2}{(R+r)m}t$

$\therefore v=u{e}^{-\frac{B^2{l}^2}{(R+r)m}t}$

Hence, $\left(\text{A}\right)$ is the correct answer.