A thin conducting rod MN of mass 20 gm, length 25 cm and resistance 10 ohm is held on frictionless,

Q: Why does the magnetic force point upwards?

By Lenz's Law, the induced current will flow in a direction that opposes the change producing it. Since the rod is falling (increasing the area and flux), the magnetic force must act upwards to oppose the downward motion.

Q: What happens if the rails had resistance?

If the rails had resistance, the total resistance $R$ of the circuit would increase as the rod falls. This would make the differential equation non-linear and much harder to solve, as $R$ would be a function of position $y$.

Q: Why is the terminal velocity constant?

As velocity increases, the opposing magnetic force increases proportionally. Eventually, this upward force exactly equals the downward gravitational force. With zero net force, acceleration becomes zero, and velocity remains constant.

Q: How did we get the exponential velocity equation?

The equation $m \frac{dv}{dt} = mg - kv$ is a standard separable differential equation. Rearranging to $\frac{dv}{mg - kv} = \frac{dt}{m}$ and integrating both sides from $v=0$ at $t=0$ yields the exponential decay form $v(t) = \frac{mg}{k}(1 - e^{-kt/m})$.

Q: Does the mass of the rod affect the terminal velocity?

Yes! Terminal velocity $v_T = \frac{mgR}{B^2 \ell^2}$ is directly proportional to mass. A heavier rod requires a larger magnetic force to balance its weight, which means it must fall faster to generate that required force.

Question

A thin conducting rod MN 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 $B_0 = 4	ext{ T}$ 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 $g = 10	ext{ ms}^{-2}$ and $e^{-1} = 0.4$]

List-I:

P. At $t = 0.2	ext{ s}$, the magnitude of the induced emf in Volt

Q. At $t = 0.2	ext{ s}$, the magnitude of the magnetic force in Newton

R. At $t = 0.2	ext{ s}$, the power dissipated as heat in Watt

S. The magnitude of terminal velocity of the rod in $	ext{m s}^{-1}$

List-II:

1. 0.07

2. 0.14

3. 1.20

4. 0.12

5. 2.00

Sigma Percentile · Accepted Answer

The problem of a conducting rod falling under gravity in a uniform magnetic field is a beautiful intersection of classical mechanics and electromagnetism. It tests our understanding of motional EMF, magnetic forces, and differential equations. Let's break down the physics step-by-step.

### Analyzing the Setup

Imagine the conducting rod MN released from rest. As it falls, gravity accelerates it downwards. However, because it is moving through a magnetic field $\vec{B}_0$ that points out of the screen, the free electrons inside the rod experience a Lorentz force. This force pushes the electrons to one end of the rod, creating a potential difference.

This phenomenon is known as **motional EMF**, given by the equation:
$$ \mathcal{E} = B_0 \ell v $$

Because the rod is in contact with perfectly conducting rails, this EMF drives an electric current $I$ through the closed loop. According to Ohm's law, the current is:
$$ I = \frac{\mathcal{E}}{R} = \frac{B_0 \ell v}{R} $$

### The Magnetic Opposing Force

Now, we have a current-carrying wire situated in a magnetic field. This means the rod will experience a magnetic force $F_m$. By Lenz's Law, the direction of this induced current will be such that the resulting magnetic force opposes the change that caused it—in this case, the downward motion.

The magnitude of this upward magnetic force is:
$$ F_m = I \ell B_0 $$

Substituting our expression for current, we get:
$$ F_m = \left(\frac{B_0 \ell v}{R}ight) \ell B_0 = \frac{B_0^2 \ell^2 v}{R} $$

Notice that this opposing force is directly proportional to the rod's velocity $v$. The faster the rod falls, the stronger the upward force becomes.

### The Master Equation

To find out how the rod moves, we apply Newton's Second Law. The net force acting on the rod is the downward gravitational force minus the upward magnetic force:
$$ F_{	ext{net}} = mg - F_m $$

$$ m \frac{dv}{dt} = mg - \frac{B_0^2 \ell^2 v}{R} $$

Dividing by mass $m$, we obtain the differential equation governing the rod's velocity:
$$ \frac{dv}{dt} = g - \left(\frac{B_0^2 \ell^2}{mR}ight) v $$

### Terminal Velocity

As the rod accelerates, its velocity increases, which in turn increases the magnetic force. Eventually, the upward magnetic force perfectly balances the downward gravitational force. At this point, the net force is zero, and the rod stops accelerating. It has reached its **terminal velocity**, $v_T$.

Setting $\frac{dv}{dt} = 0$, we solve for $v_T$:
$$ v_T = \frac{mgR}{B_0^2 \ell^2} $$

Plugging in the given values ($m = 0.02 	ext{ kg}$, $g = 10 	ext{ m/s}^2$, $R = 10 \ \Omega$, $B_0 = 4 	ext{ T}$, $\ell = 0.25 	ext{ m}$):
$$ v_T = \frac{0.02 	imes 10 	imes 10}{4^2 	imes 0.25^2} = \frac{2}{16 	imes 0.0625} = 2 	ext{ m/s} $$

### Velocity at a Specific Instant

To find the velocity at any time $t$, we solve the differential equation. Let $\alpha = \frac{B_0^2 \ell^2}{mR}$. Calculating $\alpha$:
$$ \alpha = \frac{16 	imes 0.0625}{0.02 	imes 10} = 5 	ext{ s}^{-1} $$

The solution to the differential equation is:
$$ v(t) = v_T (1 - e^{-\alpha t}) = 2(1 - e^{-5t}) $$

We need the velocity at $t = 0.2 	ext{ s}$:
$$ v(0.2) = 2(1 - e^{-5 	imes 0.2}) = 2(1 - e^{-1}) $$

Given $e^{-1} = 0.4$:
$$ v(0.2) = 2(1 - 0.4) = 1.2 	ext{ m/s} $$

### Final Calculations

With the instantaneous velocity known, we can easily compute the remaining quantities at $t = 0.2 	ext{ s}$.

**Induced EMF:**
$$ \mathcal{E} = B_0 \ell v = 4 	imes 0.25 	imes 1.2 = 1.2 	ext{ V} $$

**Magnetic Force:**
$$ F_m = \frac{B_0^2 \ell^2 v}{R} = \frac{16 	imes 0.0625 	imes 1.2}{10} = 0.12 	ext{ N} $$

**Power Dissipated:**
$$ P = \frac{\mathcal{E}^2}{R} = \frac{1.2^2}{10} = 0.144 	ext{ W} \approx 0.14 	ext{ W} $$

Matching these results with the given lists perfectly solves the matrix!

List-I

List-II

Select Matching Pairs:

Visualized Solution

\text{Physical Setup}

\text{Motional EMF}

\text{Magnetic Force}

\text{Equation of Motion}

\text{Terminal Velocity } (v_T)

\text{Velocity as a Function of Time}

\text{Velocity at } t = 0.2 \text{ s}

\text{Induced EMF at } t = 0.2 \text{ s}

\text{Magnetic Force at } t = 0.2 \text{ s}

\text{Power Dissipated at } t = 0.2 \text{ s}

\text{Final Matrix Match}

The Sigma Insight: Motional EMF

Analyzing the Setup

The Magnetic Opposing Force

The Master Equation

Terminal Velocity

Velocity at a Specific Instant

Final Calculations

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