A thin conducting rod MN of mass 20 gm, length 25 cm and resistance is held on frictionless, long, perfectly conducting vertical rails as shown in the figure. There is a uniform magnetic field directed perpendicular to the plane of the rod-rail arrangem

Question

A thin conducting rod MN of mass 20 gm, length 25 cm and resistance $10 Ω$ 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 $g = 10 {ms}^{- 2}$ and $e^{- 1} = 0.4$ ] [2023]

	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 ${ms}^{- 1}$	(4)	0.12
		(5)	2.00

Smart Achievers · Accepted Answer

(4)

From force equation, $m g - B i ℓ = m a$

$= m g - B i ℓ = m \frac{d v}{d t}$ $\Rightarrow m g - B (\frac{B i ℓ}{R}) ℓ = m \frac{d v}{d t}$ $[∵ i = \frac{ε}{R} = \frac{B ℓ v}{R}]$

$\Rightarrow \frac{m g R}{B^{2} ℓ^{2}} - v = \frac{m R}{B^{2} ℓ^{2}} \frac{d v}{d t}$

$\frac{B^{2} ℓ^{2}}{m R} \int_{0}^{t} d t = \int_{0}^{v} \frac{d v}{\frac{m g R}{B^{2} ℓ^{2}} - v}$

or, $\frac{B^{2} ℓ^{2}}{m R} = \frac{16 \times \frac{1}{16}}{20 \times 10^{- 3} \times 10} = \frac{1}{0.2} = 5$

Now, $\frac{m g R}{B^{2} ℓ^{2}} = \frac{20 \times 10^{- 3} \times 10 \times 10}{16 \times \frac{1}{16}} = 2$

And, $\frac{B^{2} ℓ^{2}}{m R} = \frac{16 \times \frac{1}{16}}{20 \times 10^{- 3} \times 10} = \frac{1}{0.2} = 5$

$∴$ $5 t = {[- \ln (2 - v)]}_{0}^{v}$ $\Rightarrow - 5 t = \ln (\frac{2 - v}{v})$

$∴$ $v = 2 (1 - e^{- 5 t})$

At $t = 0.2 s$ ,

$v = 2 (1 - e^{- 5 \times 0.2})$

$= 2 (1 - 0.4)$

$= 1.2 m/s$

At $t = 0.2 s$ , induced emf $ε = B v ℓ$

$∴$ $ε = 4 \times 1.2 \times \frac{1}{4} = 1.2 V$

Magnetic force $= B i ℓ \sin θ = B \times \frac{B ℓ v}{R} \times ℓ \times \sin 90 °$

$= \frac{4 \times 4 \times \frac{1}{4} \times 1.2 \times \frac{1}{4}}{10} = 0.12 N$

Power dissipated as heat, $P = i^{2} R - \frac{V^{2}}{R}$

$∴$ $P = \frac{1.2 \times 1.2}{10} = 0.144 W$

At terminal velocity, the net force becomes zero,

$∴$ $m g = B i ℓ$ $\Rightarrow m g = B \times \frac{B ℓ v_{t}}{R} \times ℓ$

$∴$ $v_{T} = \frac{m g R}{B^{2} ℓ^{2}} = \frac{20 \times 10^{- 3} \times 10 \times 10}{16 \times \frac{1}{16}} = 2 m/s$

Solution

1 P → 5, Q → 2, R → 3, S → 1

2 P → 3, Q → 1, R → 4, S → 5

3 P → 4, Q → 3, R → 1, S → 2

4 P → 3, Q → 4, R → 2, S → 5

Want Us to Email you About Special Offers & Updates?