A metal rod OA and mass m and length r kept rotating with a constant angular speed omega in a vertic

Q: Why do we integrate to find the EMF instead of just using $e = Bvl$?

Because the velocity of the rod isn't constant along its length! The tip is moving much faster than the pivot. We have to slice the rod into tiny pieces, find the EMF for each $dx$, and add them up using integration.

Q: Why does the current grow exponentially instead of instantly?

That's the magic of the inductor! An inductor opposes any change in current. When you close the switch, it creates a back-EMF that fights the battery, causing the current to build up slowly over time.

Q: How did we determine the direction of the magnetic force?

Using the right-hand rule. The induced current flows from A to O (inwards). The magnetic field is into the page. The cross product of current direction and magnetic field gives a force that points perpendicular to the rod, opposing the rotation.

Q: Why is there a $\cos \omega t$ term in the gravitational torque?

Torque is force times the perpendicular distance to the pivot. As the rod rotates, the horizontal distance from the center of mass to the pivot changes. It's exactly $(r/2) \cos \theta$, and since $\theta = \omega t$, we get that oscillating term.

Q: What happens to the torque when the rod is in the second quadrant?

The math handles it beautifully! When $\theta$ goes past 90 degrees, $\cos \theta$ becomes negative. This means gravity is now helping the rotation, so the external agent actually has to pull back (apply less torque) to keep the speed constant.

Question

A metal rod $OA$ and mass $m$ and length $r$ kept rotating with a constant angular speed $\omega$ in a vertical plane about horizontal axis at the end $O$. The free end $A$ is arranged to slide without friction along a fixed conducting circular ring in the same plane as that of rotation. A uniform and constant magnetic induction $\mathbf{B}$ is applied perpendicular and into the plane of rotation as shown in figure. An inductor $L$ and an external resistance $R$ are connected through a switch $S$ between the point $O$ and a point $C$ on the ring to form an electrical circuit. Neglect the resistance of the ring and the rod. Initially, the switch is open.
(a) What is the induced emf across the terminals of the switch?
(b) The switch $S$ is closed at time $t = 0$.
(i) Obtain an expression for the current as a function of time.
(ii) In the steady state, obtain the time dependence of the torque required to maintain the constant angular speed. Given that the rod $OA$ was along the positive $x$-axis at $t = 0$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in front of a giant clock, but instead of regular hands, there is a single conducting metal rod $OA$ of length $r$ and mass $m$. This rod is sweeping out a circle in a vertical plane, rotating with a constant angular speed $\omega$.

But this isn't just any empty space. The entire region is permeated by a uniform magnetic field $\mathbf{B}$ that points directly into the plane of the clock face.

As the rod slices through these invisible magnetic field lines, the free electrons inside the metal experience a Lorentz force. This continuous cutting of flux generates an electromotive force (EMF) across the ends of the rod.

Our first mission is to determine exactly how much EMF is being generated.

### The Master Equation: Motional EMF

You might be tempted to just use the standard motional EMF formula, $e = Bvl$. But there is a catch here!

The velocity of the rod isn't constant along its length. The pivot point $O$ is completely stationary ($v = 0$), while the tip $A$ is flying around at maximum speed ($v = \omega r$).

To handle this, we need the power of calculus. We slice the rod into infinitesimally small elements of length $dx$, located at a distance $x$ from the pivot.

The speed of this tiny element is simply $v = x\omega$.

The small EMF $de$ induced across this tiny element is:

$$de = Bv \, dx = B(x\omega) \, dx$$

To find the total EMF across the entire rod, we integrate this expression from the center ($x = 0$) to the tip ($x = r$):

$$e = \int_0^r B\omega x \, dx = B\omega \left[ \frac{x^2}{2} ight]_0^r$$

$$e = \frac{B\omega r^2}{2}$$

This is the constant voltage generated by our rotating rod. It acts exactly like a battery in our circuit!

### The Circuit Awakens: L-R Dynamics

Now, let's look at the rest of the setup. The pivot $O$ and the outer conducting ring are connected to an external circuit containing a resistor $R$, an inductor $L$, and a switch $S$.

When we close the switch at $t = 0$, we are essentially connecting our "rod-battery" to an $L-R$ circuit.

Does the current instantly jump to its maximum value? Absolutely not!

The inductor $L$ acts like electrical inertia. It despises sudden changes in current and creates a back-EMF to fight the rising current.

The current grows exponentially according to the classic $L-R$ growth equation:

$$i(t) = i_0 \left(1 - e^{-t/	au_L}ight)$$

Here, the steady-state maximum current is $i_0 = \frac{e}{R}$, and the time constant of the circuit is $	au_L = \frac{L}{R}$.

Substituting our calculated EMF into this equation, we get the complete expression for the current as a function of time:

$$i(t) = \frac{B\omega r^2}{2R} \left[ 1 - e^{-(R/L)t} ight]$$

This beautifully describes how the current slowly builds up its strength against the inductor's resistance.

### The Steady State and The Battle of Torques

Let's fast forward in time. After a long time ($t 	o \infty$), the exponential term $e^{-(R/L)t}$ decays to zero. The inductor finally yields, and the current reaches its steady, maximum value:

$$i = \frac{B\omega r^2}{2R}$$

Now, the physics gets incredibly dynamic. We have a steady current flowing through a rod that is moving in a magnetic field.

By the right-hand rule, the induced current flows inwards from $A$ to $O$. A current-carrying wire in a magnetic field experiences a magnetic force $F_m = i r B$.

According to Lenz's Law, this force must oppose the cause that produced it. Since the rod is rotating counter-clockwise, the magnetic force pushes back in the **clockwise** direction.

This force effectively acts at the center of mass of the rod, which is at a distance of $r/2$ from the pivot. The torque generated by this magnetic drag is:

$$	au_m = F_m \cdot \frac{r}{2} = (i r B) \frac{r}{2} = \left( \frac{B\omega r^2}{2R} ight) \frac{r^2 B}{2}$$

$$	au_m = \frac{B^2\omega r^4}{4R} \quad 	ext{(clockwise)}$$

But wait, there is another player on the field: **Gravity!**

The rod has a mass $m$, and its weight $mg$ pulls straight down from the center of mass.

The torque due to gravity depends on the angle $	heta$ of the rod. The perpendicular distance from the pivot to the line of action of gravity is $\frac{r}{2} \cos	heta$.

Since the rod rotates with constant angular speed $\omega$, the angle at any time $t$ is $	heta = \omega t$.

$$	au_g = (mg) \left( \frac{r}{2} \cos	heta ight) = \frac{mgr}{2} \cos\omega t \quad 	ext{(clockwise in 1st quadrant)}$$

### The Final Balance

We are told that the rod maintains a **constant** angular speed $\omega$.

Newton's Second Law for rotation tells us that if the angular acceleration is zero, the net torque must be zero!

This means some external agent (maybe a motor) must be constantly applying a torque to fight against the magnetic drag and the fluctuating pull of gravity.

To keep the rod spinning smoothly, the external torque must exactly balance the sum of the opposing torques:

$$	au_{	ext{ext}} = 	au_m + 	au_g$$

$$	au_{	ext{ext}} = \frac{B^2\omega r^4}{4R} + \frac{mgr}{2} \cos\omega t$$

This final equation is a masterpiece. It shows a constant term representing the relentless electromagnetic braking, combined with a sinusoidal term representing the rhythmic rise and fall of the gravitational pull as the rod sweeps through its circular path.

Physics doesn't get much more elegant than this!

Visualized Solution

\text{Analyzing the Setup}

\text{Motional EMF in a Rotating Rod}

\text{Total Induced EMF}

\text{Equivalent Circuit}

\text{Current Growth in } L-R \text{ Circuit}

\text{Current as a Function of Time}

\text{Steady State Analysis}

\text{Magnetic Force on the Rod}

\text{Torque due to Magnetic Force}

\text{Torque due to Gravity}

\text{Net External Torque Required}

\text{Final Conclusion}

The Sigma Insight: Motional EMF

Analyzing the Setup

The Master Equation

The Circuit Awakens

The Steady State and The Battle of Torques

The Final Balance

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