A thin conducting rod M N of mass 20 ~gm , length 25 ~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 ~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 ~m ~s^-2 and e^-1=0.4 ] .tg border-collapse:collapse;border-spacing:0;width:100% .tg tdborder-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal; .tg thborder-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal; .tg .tg-1wigfont-weight:bold;text-align:left;vertical-align:top .tg .tg-0laxtext-align:left;vertical-align:top List - I List - II (P) At t=0.2 ~s , the magnitude of the induced emf in Volt (1) 0.07 (Q) At t=0.2 ~s , the magnitude of the magnetic force in Newton (2) 0.14 (R) At t=0.2 ~s , the power dissipated as heat in Watt (3) 1.20 (S) The magnitude of terminal velocity of the rod in m s^-1 (4) 0.12 (5) 2.00

P arrow 3, Q arrow 4, R arrow 2, S arrow 5

P arrow 5, Q arrow 2, R arrow 3, S arrow 1

P arrow 3, Q arrow 1, R arrow 4, S arrow 5

P arrow 4, Q arrow 3, R arrow 1, S arrow 2

Electromagnetic Induction PYQ — JEE Advanced 2023 · Shift 1 · Q51 with Solution

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

]

JEE Advanced 2023 Paper 1 Online Physics - Electromagnetic Induction Question 4 English

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

A
$P \rightarrow 5, Q \rightarrow 2, R \rightarrow 3, S \rightarrow 1$
B
$P \rightarrow 3, Q \rightarrow 1, R \rightarrow 4, S \rightarrow 5$
C
$P \rightarrow 4, Q \rightarrow 3, R \rightarrow 1, S \rightarrow 2$
D
$P \rightarrow 3, Q \rightarrow 4, R \rightarrow 2, S \rightarrow 5$

View written solutionFree

Correct answer: D

Set up the equation of motion

When the rod moves downward with speed $v$ , the induced emf is

$\mathcal{E}=B\ell v$

where

$B=4\,\text{T}$
$\ell=25\,\text{cm}=0.25\,\text{m}$
$R=10\,\Omega$
$m=20\,\text{g}=0.02\,\text{kg}$

So current in the circuit is

$I=\frac{\mathcal{E}}{R}=\frac{B\ell v}{R}$

Magnetic force on the rod is upward:

$F_B=B I \ell = B\ell\left(\frac{B\ell v}{R}\right)=\frac{B^2\ell^2}{R}v$

Equation of motion downward:

$m\frac{dv}{dt}=mg-\frac{B^2\ell^2}{R}v$

This is of the form

$\frac{dv}{dt}+\frac{B^2\ell^2}{mR}v=g$

Its solution is

$v(t)=v_t\left(1-e^{-t/\tau}\right)$

where terminal velocity

$v_t=\frac{mgR}{B^2\ell^2}$

and time constant

$\tau=\frac{mR}{B^2\ell^2}$

Compute terminal velocity

First,

$B^2\ell^2=4^2\times (0.25)^2=16\times 0.0625=1$

Hence

$v_t=\frac{mgR}{B^2\ell^2}=\frac{0.02\times 10\times 10}{1}=2.0\,\text{m s}^{-1}$

$S \rightarrow 5$

Compute time constant

$\tau=\frac{mR}{B^2\ell^2}=\frac{0.02\times 10}{1}=0.2\,\text{s}$

Thus at $t=0.2\,$ s,

$\frac{t}{\tau}=1$

$v(0.2)=v_t(1-e^{-1})=2(1-0.4)=2(0.6)=1.2\,\text{m s}^{-1}$

Find induced emf at $t=0.2$ s

$\mathcal{E}=B\ell v = 4\times 0.25\times 1.2$

Since $4\times 0.25=1$ ,

$\mathcal{E}=1.2\,\text{V}$

$P \rightarrow 3$

Find magnetic force at $t=0.2$ s

Current:

$I=\frac{\mathcal{E}}{R}=\frac{1.2}{10}=0.12\,\text{A}$

Then magnetic force:

$F_B=B I \ell =4\times 0.12\times 0.25$

Again $4\times 0.25=1$ , so

$F_B=0.12\,\text{N}$

$Q \rightarrow 4$

Find power dissipated as heat at $t=0.2$ s

$P_{\text{heat}}=I^2R=(0.12)^2\times 10=0.0144\times 10=0.144\,\text{W}$

Thus approximately

$P_{\text{heat}}\approx 0.14\,\text{W}$

$R \rightarrow 2$

Final matching

We get:

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

This corresponds to Option D.

More from Electromagnetic Induction

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

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

A
$P \rightarrow 5, Q \rightarrow 2, R \rightarrow 3, S \rightarrow 1$
B
$P \rightarrow 3, Q \rightarrow 1, R \rightarrow 4, S \rightarrow 5$
C
$P \rightarrow 4, Q \rightarrow 3, R \rightarrow 1, S \rightarrow 2$
D
$P \rightarrow 3, Q \rightarrow 4, R \rightarrow 2, S \rightarrow 5$

View written solutionFree

Correct answer: D

Set up the equation of motion

When the rod moves downward with speed $v$ , the induced emf is

$\mathcal{E}=B\ell v$

where

$B=4\,\text{T}$
$\ell=25\,\text{cm}=0.25\,\text{m}$
$R=10\,\Omega$
$m=20\,\text{g}=0.02\,\text{kg}$

So current in the circuit is

$I=\frac{\mathcal{E}}{R}=\frac{B\ell v}{R}$

Magnetic force on the rod is upward:

$F_B=B I \ell = B\ell\left(\frac{B\ell v}{R}\right)=\frac{B^2\ell^2}{R}v$

Equation of motion downward:

$m\frac{dv}{dt}=mg-\frac{B^2\ell^2}{R}v$

This is of the form

$\frac{dv}{dt}+\frac{B^2\ell^2}{mR}v=g$

Its solution is

$v(t)=v_t\left(1-e^{-t/\tau}\right)$

where terminal velocity

$v_t=\frac{mgR}{B^2\ell^2}$

and time constant

$\tau=\frac{mR}{B^2\ell^2}$

Compute terminal velocity

First,

$B^2\ell^2=4^2\times (0.25)^2=16\times 0.0625=1$

Hence

$v_t=\frac{mgR}{B^2\ell^2}=\frac{0.02\times 10\times 10}{1}=2.0\,\text{m s}^{-1}$

$S \rightarrow 5$

Compute time constant

$\tau=\frac{mR}{B^2\ell^2}=\frac{0.02\times 10}{1}=0.2\,\text{s}$

Thus at $t=0.2\,$ s,

$\frac{t}{\tau}=1$

$v(0.2)=v_t(1-e^{-1})=2(1-0.4)=2(0.6)=1.2\,\text{m s}^{-1}$

Find induced emf at $t=0.2$ s

$\mathcal{E}=B\ell v = 4\times 0.25\times 1.2$

Since $4\times 0.25=1$ ,

$\mathcal{E}=1.2\,\text{V}$

$P \rightarrow 3$

Find magnetic force at $t=0.2$ s

Current:

$I=\frac{\mathcal{E}}{R}=\frac{1.2}{10}=0.12\,\text{A}$

Then magnetic force:

$F_B=B I \ell =4\times 0.12\times 0.25$

Again $4\times 0.25=1$ , so

$F_B=0.12\,\text{N}$

$Q \rightarrow 4$

Find power dissipated as heat at $t=0.2$ s

$P_{\text{heat}}=I^2R=(0.12)^2\times 10=0.0144\times 10=0.144\,\text{W}$

Thus approximately

$P_{\text{heat}}\approx 0.14\,\text{W}$

$R \rightarrow 2$

Final matching

We get:

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

This corresponds to Option D.

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