Question

A metal rod $\text{OA}$ of length $l=50\text{cm}$ is kept rotating with a constant angular speed of $\omega =10{\text{rad s}}^{-1}$ about point $\text{O}$ as shown in the figure. The free end $\text{A}$ is arranged to slide without friction along a fixed uniform circular ring in the same plane as that of rotation. A uniform magnetic field of magnitude $B=2\text{T}$ exists in the region as shown in the figure. An inductor of inductance $L=4\text{H}$ and an external resistance $R=8\Omega$ are connected through a switch $\text{S}.$ If the switch is closed at $t=0$, find the current flowing through the resistance at $t=1\text{s}.$

[Use $e^{-2}=0.135$]

• A. $2\text{A}$
• B. $1.5\text{A}$
• C. $0.27\text{A}$
• D. $0.08\text{A}$

Answer 1

The correct option is C $0.27\text{A}$

Given:
$l=50\text{cm};\omega =10{\text{rad s}}^{-1};t=1\text{s}$
$L=4\text{H};B=2\text{T};R=8\Omega$


Due to rotation of the rod $\text{OA}$ with a constant angular speed $\omega$ in a uniform magnetic field, an emf $E$ induced across the rod $\text{OA}$ is,

$E=\frac{B\omega l^2}{2}=\frac{2\times 10\times \frac{1}{4}}{2}=2.5\text{V}$

So the rod will behave as a battery of emf $2.5\text{V}.$

$\therefore$ The current flowing in the given circuit at any instant time $t$ is,

$i=i_0(1-e^{-t/\tau }).....\left(1\right)$

$i=\frac{E}{R}(1-e^{-t/\tau })$

The time constant of the given circuit is,

$\tau =\frac{L}{R}=\frac{4}{8}=\frac{1}{2}\text{s}$

Putting this value in (1) we get,

$i=\frac{2.5}{8}(1-e^{-t/\left(1/2\right)})=\frac{2.5}{8}(1-e^{-2t})$

At time, $t=1\text{s}$

$i=\frac{2.5}{8}(1-e^{-2})$

$\therefore i=0.27\text{A}$

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

Why this question ? This question is an example of how different concepts can be merged together.