A pair of parallel horizontal conducting rails of negligible resistance shorted at one end is fixed

Q: Why is the net force on the rod zero even when it's accelerating?

Because the rod is explicitly stated to be *massless* ($m_{rod} = 0$). According to Newton's second law ($F_{net} = m_{rod}a$), if the mass is zero, the net force must be zero regardless of the acceleration. This means the tension $T$ pulling it forward must perfectly equal the magnetic force $F_m$ pulling it backward.

Q: How do we know the direction of the induced current?

You can use Lenz's Law or the right-hand rule. As the rod moves right, the area of the loop increases, increasing the outward magnetic flux. To oppose this, the induced current must create an inward magnetic field, which means it flows clockwise. Alternatively, the force on positive charges $\vec{v} \times \vec{B}$ points downwards along the rod.

Q: Why does the acceleration decrease over time?

As the rod speeds up, it cuts through the magnetic field faster, which increases the induced EMF and current. This larger current results in a stronger opposing magnetic force. Since gravity pulling the mass down is constant, the net force (and thus acceleration) decreases as the opposing magnetic force grows.

Q: What exactly is terminal velocity in this context?

Terminal velocity is the maximum, constant speed the system reaches. It occurs when the opposing magnetic force grows large enough to perfectly balance the gravitational force on the hanging mass. At this point, the net force is zero, so the system stops accelerating and moves at a steady speed.

Q: Why doesn't the resistance of the rails matter?

The problem states that the rails have 'negligible resistance'. This is a common idealization in physics problems to simplify the math. If the rails had resistance, the total resistance of the loop would increase as the rod moves, making the current and magnetic force dependent on position, not just velocity, leading to a much more complex differential equation!

Question

A pair of parallel horizontal conducting rails of negligible resistance shorted at one end is fixed on a table. The distance between the rails is $L$. A conducting massless rod of resistance $R$ can slide on the rails frictionlessly. The rod is tied to a massless string which passes over a pulley fixed to the edge of the table. A mass $m$ tied to the other end of the string hangs vertically. A constant magnetic field $B$ exists perpendicular to the table. If the system is released from rest. Calculate :
(a) the terminal velocity achieved by the rod, and
(b) the acceleration of the mass at the instant when the velocity of the rod is half the terminal velocity.

Sigma Percentile · Accepted Answer

### The Setup: A Dance of Mechanics and Electromagnetism

Imagine a perfectly smooth table with two parallel conducting rails separated by a distance $L$. A massless conducting rod of resistance $R$ rests across these rails. This rod is connected via a massless string over a pulley to a hanging mass $m$. The entire setup is immersed in a uniform magnetic field $B$ pointing perpendicular to the table.

When we release the system from rest, gravity pulls the mass $m$ downwards. This creates tension in the string, which in turn pulls the rod along the rails. As the rod begins to move, it enters the fascinating realm of electromagnetic induction.

### The Birth of Motional EMF

As the rod slides with a velocity $v$, it cuts through the perpendicular magnetic field lines. According to Faraday's Law of Induction, this motion generates a motional electromotive force (EMF) across the ends of the rod. The magnitude of this induced EMF is given by the elegant equation:

$$e = BvL$$

Because the rails are shorted at one end, they form a closed electrical loop with the rod. This induced EMF acts like a battery, driving a current through the loop. By Ohm's law, the induced current $i$ is simply the EMF divided by the resistance $R$ of the rod (since the rails have negligible resistance):

$$i = \frac{e}{R} = \frac{BvL}{R}$$

### The Opposing Force: Lenz's Law in Action

Nature loves balance. According to Lenz's Law, the induced current will flow in a direction that opposes the change causing it. In this case, the current-carrying rod is moving through a magnetic field, so it experiences a magnetic Lorentz force. The magnitude of this force is:

$$F_m = iLB$$

Substituting our expression for the current, we get:

$$F_m = \left(\frac{BvL}{R}\right)LB = \frac{B^2L^2v}{R}$$

This magnetic force acts in the direction opposite to the rod's velocity, acting like an electromagnetic drag or friction that grows stronger as the rod speeds up.

### Newton's Laws: The Master Equation

Now, let's bridge the gap between electromagnetism and classical mechanics using Newton's second law.

First, consider the rod. The problem explicitly states that the rod is **massless** ($m_{rod} = 0$). Therefore, the net force on it must be zero. The tension $T$ pulling it forward must perfectly balance the magnetic force $F_m$ pulling it backward:

$$T - F_m = 0 \implies T = F_m$$

Next, consider the hanging mass $m$. Gravity pulls it down with a force $mg$, while the tension $T$ pulls it up. Its equation of motion is:

$$mg - T = ma$$

Substituting $T = F_m$ into this equation, we get our master equation for the system:

$$mg - F_m = ma$$

$$mg - \frac{B^2L^2v}{R} = ma$$

Dividing by $m$, we find the acceleration as a function of velocity:

$$a = g - \frac{B^2L^2v}{mR}$$

### Reaching the Limit: Terminal Velocity

As the system accelerates, the velocity $v$ increases. Consequently, the opposing magnetic force $\frac{B^2L^2v}{R}$ also increases. Eventually, this opposing force becomes exactly equal to the gravitational force $mg$.

At this precise moment, the net force on the system is zero, and the acceleration $a$ drops to zero. The rod stops accelerating and continues to move at a constant maximum speed known as the **terminal velocity** ($v_T$).

Setting $a = 0$ in our acceleration equation:

$$0 = g - \frac{B^2L^2v_T}{mR}$$

Solving for $v_T$, we get the answer to the first part of our problem:

$$v_T = \frac{mgR}{B^2L^2}$$

### The Halfway Point: A Beautiful Cancellation

The second part of the question asks for the acceleration of the mass when the rod's velocity is exactly half of its terminal velocity.

Let's substitute $v = \frac{v_T}{2}$ into our acceleration equation:

$$v = \frac{1}{2} \left( \frac{mgR}{B^2L^2} \right)$$

$$a = g - \frac{B^2L^2}{mR} \left( \frac{mgR}{2B^2L^2} \right)$$

Watch how beautifully the terms cancel out! The $B^2L^2$ in the numerator cancels with the denominator, and the $mR$ terms cancel as well. We are left with:

$$a = g - \frac{g}{2} = \frac{g}{2}$$

At exactly half the terminal velocity, the opposing magnetic force is exactly half the weight of the hanging mass. Therefore, the net force is half the weight, and the acceleration is exactly half of the acceleration due to gravity.

This problem is a stunning showcase of how mechanical forces and electromagnetic induction intertwine to create a self-regulating dynamic system!

Visualized Solution

\text{System Overview}

\text{Motional EMF}

\text{Induced Current}

\text{Magnetic Force}

\text{Equations of Motion}

\text{Acceleration as a function of velocity}

\text{Terminal Velocity } (v_T)

\text{Acceleration at } v = \frac{v_T}{2}

\text{Final Calculation}

The Sigma Insight: Motional EMF

The Setup

The Birth of Motional EMF

The Opposing Force

Newton's Laws

Reaching the Limit

The Halfway Point

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