Q.A $1.0\ \text{m}$ long metallic rod is rotated with an angular frequency of $400\ \text{rad s}^{-1}$ about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of $0.5\ \text{T}$ parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.

The key idea is motional emf in a rotating rod: each segment of the rod moves with a different linear speed, so we integrate the motional emf $d\mathcal{E} = B\,v\,dr$ along the length.

1. For a rod of length $L = 1.0\ \text{m}$ rotating at $\omega = 400\ \text{rad/s}$ about one end, the linear speed at a distance $r$ from the centre is $v = \omega r$.
2. The motional emf across a small element $dr$ is $d\mathcal{E} = B\,v\,dr = B\omega r\,dr$.
3. Integrate from $r = 0$ (centre) to $r = L$ (ring end): $\mathcal{E} = \int_0^L B\omega r\,dr = B\omega \frac{L^2}{2}.$ …

The key idea here is motional emf — an emf is induced when a conductor moves through a magnetic field, because the free charges inside experience a magnetic Lorentz force that pushes them along the rod. When the rod rotates, every tiny segment of it is moving with a different speed (faster near the tip, slower near the pivot), so we cannot just use $Blv$ with a single $v$. Instead, we integrate the contribution from each infinitesimal piece.

Think of it this way: the rod is like a radial spoke sweeping across the magnetic field. Each point on the rod has a velocity perpendicular to both the rod and the field, so the motional emf builds up from the centre outward. The ring simply provides a path to collect that emf — the emf between the centre and the ring is exactly the same as the emf generated along the whole rod.

Let’s work through it step by step.

1. Set up the geometry and the physics

   The rod is $L = 1.0\ \text{m}$ long, rotating at $\omega = 400\ \text{rad/s}$ about one end. The magnetic field $B = 0.5\ \text{T}$ is parallel to the rotation axis — that means it is perpendicular to the plane of rotation.

   For a small element of the rod at a distance $r$ from the centre (pivot), its linear speed is $v = \omega r$. The direction of motion is tangential, which is perpendicular to both the rod (radial) and the field (axial). So the motional emf across that element is $d\mathcal{E} = B\, v\, dr = B \omega r\, dr$.

2. Integrate along the rod

   The emf from the centre ($r=0$) to the tip ($r=L$) is the sum of all these tiny contributions:

   $\mathcal{E} = \int_0^L B \omega r\, dr = B\omega \int_0^L r\, dr = B\omega \cdot \frac{L^2}{2}$ …

Method: Motional EMF in a Rotating Rod (Using Faraday's Law / Induced Electric Field)

This is a classic problem where a conductor rotates in a uniform magnetic field. The key insight: different parts of the rod move at different speeds, so we cannot simply use $\mathcal{E} = Blv$.

Steps

Step 1: Understand the setup

• Rod length: $l = 1.0\ \text{m}$
• Angular speed: $\omega = 400\ \text{rad/s}$
• Magnetic field: $B = 0.5\ \text{T}$, parallel to the axis of rotation
• The rod rotates about one end; the other end slides on a circular ring.

Step 2: Consider a small element of the rod

Take a small segment of the rod at a distance $x$ from the centre (pivot). Its linear velocity is:

Step 3: Motional emf across the small element

For a conductor moving perpendicular to $B$, the motional emf across a small length $dx$ is:

Step 4: Integrate over the entire rod

The total emf between the centre ($x=0$) and the ring ($x=l$) is:

$\mathcal{E} = \int_0^l B\omega x\,dx = B\omega \int_0^l x\,dx$ …

Here are the most common mistakes students make when solving this classic Motional EMF problem — and how to avoid each one.

✗ Mistake 1: Using the wrong formula for motional emf

Many students directly write

$\mathcal{E} = B l v$

and then plug in $v = \omega l$, getting

$\mathcal{E} = B l (\omega l) = B \omega l^2$.

Why it’s wrong:

The formula $\mathcal{E} = B l v$ works only when the entire rod moves with the same velocity perpendicular to the field. Here, different points on the rod have different linear speeds ($v = \omega r$). The emf is not simply $B \omega l^2$.

How to avoid:

Always derive from first principles:

• For a rotating rod, the emf is the average of $B v$ over the length.
• Use $\mathcal{E} = \int_0^l B \, \omega r \, dr = \frac{1}{2} B \omega l^2$.

✓ Correct formula:

✗ Mistake 2: Forgetting the factor of $\frac{1}{2}$

Even if students set up the integral, they sometimes forget the $\frac{1}{2}$ and write $\mathcal{E} = B \omega l^2$.

Why it’s wrong:

The integral $\int_0^l r \, dr = \frac{l^2}{2}$, not $l^2$.

How to avoid:

• Always write the integral explicitly:

• Never skip steps — the $\frac{1}{2}$ is the most common point of error.

✗ Mistake 3: Confusing the axis of rotation

Students sometimes think the rod rotates about its centre instead of one end.

Why it’s wrong:

• If rotated about the centre, the limits of integration change ($-l/2$ to $l/2$), and the net emf becomes zero (symmetry).
• Here, rotation is about one end, so the rod sweeps a full radius $l$.

How to avoid:

• Read carefully: “passing through its one end” means the axis is at one end.
• Draw a diagram — mark the pivot point and the ring contact at the other end.

✗ Mistake 4: Mixing up which end is connected to the ring

Some students think the emf is measured between the two ends of the rod.

Why it’s wrong:

The question asks for emf between the centre (pivot) and the ring (other end). That’s exactly the potential difference across the whole rod.

How to avoid:

• The centre is at $r=0$, the ring at $r=l$. …