Animated Solution for Physics - Electromagnetic Induction: A metallic rod of length $l$ is tied to a string of length $2l$ and made to rotate with angular speed $\omega$ on a horizontal table with one end of the string fixed. If there is a vertical magnetic field $B$ in the region, the emf induced across the ends of the rod is

A Merry-Go-Round of Electrons

Imagine you are looking down at a horizontal table. There is a uniform magnetic field $\vec{B}$ piercing straight down into the surface. At the center of the table, a string of length $2l$ is firmly anchored. Attached to the other end of this string is a metallic rod of length $l$.

Suddenly, the entire system starts spinning like a merry-go-round with a constant angular velocity $\omega$. As the metallic rod sweeps through the magnetic field, the free electrons inside it experience a magnetic Lorentz force. This force pushes the electrons to one end of the rod, creating a potential difference. Our goal is to find the exact value of this induced electromotive force (EMF).

Why the Standard Formula Fails

You might be tempted to jump straight to the standard motional EMF formula, $e=Bvl$. However, there is a massive catch here.

The formula $e=Bvl$ only works when the entire conductor is moving at a constant, uniform velocity. But in our rotating system, the rod is not moving uniformly! The inner tip of the rod (attached to the string) is moving slower than the outer tip. Because the velocity $v$ depends on the distance from the center ($v=\omega x$), we cannot use a single value for $v$.

To solve this, we must summon the power of calculus.

Slicing the Rod

Instead of looking at the whole rod, let's zoom in and consider an infinitesimally small slice of the rod, which we will call $dx$. Let's say this tiny slice is located at a distance $x$ from the central pivot point.

Because this slice is so incredibly small, we can assume that all parts of it are moving at the exact same velocity. From the kinematics of circular motion, we know that the linear velocity of this slice is:

Now, we can safely apply Faraday's law of motional EMF to this tiny slice. The small EMF $de$ induced across our element $dx$ is:

Substituting our expression for velocity, we get:

Adding it All Up

We have the EMF for a tiny slice, but we want the EMF for the entire rod. To get the total EMF, we must integrate $de$ over the entire length of the rod.

This is where many students make a critical mistake. What should the limits of integration be?

The rod does not start at the center! The space from $x=0$ to $x=2l$ is occupied by the non-conducting string. The metallic rod only begins at a distance of $2l$ from the pivot. Since the rod has a length of $l$, it ends at a distance of $2l+l=3l$.

Therefore, our limits of integration are strictly from $x=2l$ to $x=3l$. Let's set up the integral:

Since the magnetic field $B$ and the angular velocity $\omega$ are constants, we can pull them outside the integral:

The integral of $x$ with respect to $x$ is simply $\frac{x^2}{2}$. Let's evaluate this with our limits:

And there we have it! The total induced EMF across the ends of the rod is $\frac{5B\omega l^2}{2}$.

This problem beautifully demonstrates why memorizing formulas isn't enough in physics. You must understand the physical reality behind the equations. Always pay attention to the axis of rotation and the physical boundaries of your conductors!