Question 30: Electromagnetic Induction

Question

A thin conducting rod $M N$ of mass $20 ~\mathrm{gm}$, length $25 \mathrm{~cm}$ and resistance $10 ~\Omega$ is held on frictionless, long, perfectly conducting vertical rails as shown in the figure. There is a uniform magnetic field $B_0=4 \mathrm{~T}$ directed perpendicular to the plane of the rod-rail arrangement. The rod is released from rest at time $t=0$ and it moves down along the rails. Assume air drag is negligible. Match each quantity in List-I with an appropriate value from List-II, and choose the correct option.
[Given: The acceleration due to gravity $g=10 \mathrm{~m} \mathrm{~s}^{-2}$ and $e^{-1}=0.4$ ]

List - I
List - II

(P) At $t=0.2 \mathrm{~s}$, the magnitude of the induced emf in Volt
(1) 0.07

(Q) At $t=0.2 \mathrm{~s}$, the magnitude of the magnetic force in Newton
(2) 0.14

(R) At $t=0.2 \mathrm{~s}$, the power dissipated as heat in Watt
(3) 1.20

(S) The magnitude of terminal velocity of the rod in $\mathrm{m} \mathrm{s}^{-1}$
(4) 0.12

(5) 2.00

Accepted Answer

From force equation
$$
\begin{aligned}
& \mathrm{mg}-\mathrm{Bi} \ell=\frac{\mathrm{mdv}}{\mathrm{dt}} \\
& \mathrm{mg}-\frac{\mathrm{BBi} \ell}{\mathrm{R}} 	imes \ell=\frac{\mathrm{mdv}}{\mathrm{dt}} \\
& \frac{\mathrm{mgR}}{\mathrm{B}^2 \ell^2}-\mathrm{v}=\frac{\mathrm{mR}}{\mathrm{B}^2 \ell^2} \frac{\mathrm{dv}}{\mathrm{dt}} \\
& \frac{\mathrm{B}^2 \ell^2}{\mathrm{mR}} \int\limits_{\mathrm{t}=0}^{\mathrm{t}} \mathrm{dt}=\int\limits_0^{\mathrm{v}} \frac{\mathrm{dv}}{\left(\frac{\mathrm{mgR}}{\mathrm{B}^2 \ell^2}-\mathrm{v}ight)}
\end{aligned}
$$
Now $\frac{\mathrm{mgR}}{\mathrm{B}^2 \ell^2}=\frac{20 	imes 10^{-3} 	imes 10 	imes 10}{16 	imes \frac{1}{16}}=2$
And $\frac{\mathrm{B}^2 \ell^2}{\mathrm{mR}}=\frac{16 	imes \frac{1}{16}}{20 	imes 10^{-3} 	imes 10}=\frac{1}{0.2}=5$
$\begin{aligned} & 	herefore 5 \mathrm{t}=[-\ln (2-\mathrm{v})]_0^{\mathrm{v}} \\ & -5 \mathrm{t}=\ell \mathrm{n}\left[\frac{2-\mathrm{v}}{\mathrm{v}}ight] \\ & 	herefore \mathrm{v}=2\left(1-\mathrm{e}^{-5 \mathrm{t}}ight)\end{aligned}$
At $\mathrm{t}=0.2 \mathrm{sec}$
$\begin{aligned} & \mathrm{v}=2\left(1-\mathrm{e}^{-5 	imes 0.2}ight) \\ & \mathrm{v}=2(1-0.4) \\ & \mathrm{v}=1.2 \mathrm{~m} / \mathrm{s}\end{aligned}$
(P) Now at $\mathrm{t}=0.2 \mathrm{sec}$
The magnitude of the induced emf $=\mathrm{E}=\mathrm{Bv} \ell$
$$
=4 	imes 1.2 	imes \frac{1}{4}=1.2 \mathrm{Volt}
$$
(Q) At $\mathrm{t}=0.2 \mathrm{sec}$, the magnitude of magnetic force $=\mathrm{BI} \ell \sin 	heta$
$$
\begin{aligned}
& =\mathrm{B} 	imes \frac{\mathrm{B} \ell \mathrm{v}}{\mathrm{R}} 	imes \ell 	imes \sin 90^{\circ} \\
& =\frac{4 	imes 4 	imes \frac{1}{4} 	imes 1.3 	imes \frac{1}{4}}{10}=0.12 	ext { Newton }
\end{aligned}
$$
(R) At t $=0.2 \mathrm{sec}$, the power dissipated as heat
$$
\begin{aligned}
& P=i^2 R=\frac{v^2}{R}=\frac{1.2 	imes 1.2}{10} \\
& P=0.144 	ext { watt }
\end{aligned}
$$
(S) Magnitude of terminal velocity
At terminal velocity, the net force become zero
$$
\begin{aligned}
& 	herefore \mathrm{mg}=\mathrm{Bi} \ell \\
& \mathrm{mg}=\mathrm{B} 	imes \frac{\mathrm{B} \ell \mathrm{v}_{\mathrm{t}}}{\mathrm{R}} 	imes \ell \\
& 	herefore \mathrm{v}_{\mathrm{T}}=\frac{\mathrm{mgR}}{\mathrm{B}^2 \ell^2}=\frac{20 	imes 10^{-3} 	imes 10 	imes 10}{16 	imes \frac{1}{16}} \\
& \mathrm{v}_{\mathrm{T}}=2 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
Hence, Answer is (D)

Answer

$P \rightarrow 5, Q \rightarrow 2, R \rightarrow 3, S \rightarrow 1$

Answer

$P \rightarrow 3, Q \rightarrow 1, R \rightarrow 4, S \rightarrow 5$

Answer

$P \rightarrow 4, Q \rightarrow 3, R \rightarrow 1, S \rightarrow 2$

Answer

$P \rightarrow 3, Q \rightarrow 4, R \rightarrow 2, S \rightarrow 5$

Electromagnetic Induction

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