Question

Comprehension:

A conducting rod of mass $m$ and length $l$ is released from rest on a smooth metallic rails placed in vertical plane in a uniform horizontal magnetic field $\left(B\right)$ as shown in the figure. When rod falls, it would cut magnetic field lines and motional emf will be induced. Velocity and acceleration of rod will change with time and after a long time rod will achieve a maximum velocity named as terminal velocity.


Answer the following question based on above discussion.

$\left(iv\right)$ Which of the options is true for the above passage ?

• A. When the rod has achieved terminal velocity, current in the resistor is zero.
• B. Current in the resister first increases, then decrease.
• C. The acceleration - velocity graph of the motion of the rod will be a straight line.
• D. The velocity-time graph of rod will be a parabola.

Answer 1

The correct option is C The acceleration - velocity graph of the motion of the rod will be a straight line.

(i). When the rod achieves terminal velocity, motional emf will be constant, so current in the resistor will be non-zero and constant.
$\Rightarrow$Option $\left(A\right)$ is incorrect.

(ii). Velocity of rod continuously increase and emf $\left(Bvl\right)$ will also increase till terminal velocity is achieved, after that it will become constant
$\Rightarrow$Option $\left(B\right)$ is incorrect.

(iii). Acceleration-velocity graph of the rod is,

At any instant , $t>0$, from the FBD of the rod, we get,


$mg-\frac{B^2{l}^{2}v}{R}=ma.......\left(1\right)\left[\text{this is similar to the equation of straight line}\right]$


$\Rightarrow$Option $\left(C\right)$ is correct.

(iv). From (1) discussed above,

$\frac{dv}{dt}=g-\frac{B^2{l}^{2}v}{mR}$

$\Rightarrow {\int }_0^v\frac{dv}{(g-\frac{B^2{l}^{2}v}{mR})}={\int }_0^{t}dt$

The above equation converts to exponential form.
$\Rightarrow$Option $\left(D\right)$ is incorrect.

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.