Q.A metallic rod of $1\ \text{m}$ length is rotated with a frequency of $50\ \text{rev/s}$, with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius $1\ \text{m}$, about an axis passing through the centre and perpendicular to the plane of the ring (Fig. 6.11).

The key idea is motional emf in a rotating rod: each segment of the rod cuts magnetic field lines at a different speed, so the emf is found by integrating $d\mathcal{E} = B\,v\,dr$ along the rod.

1. The rod rotates with angular velocity $\omega = 2\pi f = 2\pi \times 50 = 100\pi\ \text{rad/s}$.
2. At a distance $r$ from the centre, the linear speed is $v = \omega r$. The motional emf across a small element $dr$ is $d\mathcal{E} = B\,v\,dr = B\omega r\,dr$.
3. Integrate from $r=0$ to $r=R=1\ \text{m}$:

4. Substitute values: $B=1\ \text{T}$, $\omega=100\pi\ \text{rad/s}$, $R=1\ \text{m}$:

Step-by-Step Solution

Take a small element at distance $x$ from the hinge; it moves with speed $v=\omega x$. The motional emf across the whole rod is

Angular frequency:

Therefore:

Method: Motional EMF in a Rotating Rod (Faraday's Law in Rotating Frame)

This is a classic problem of a rotating conductor in a uniform magnetic field. The key insight: different parts of the rod move at different speeds, so we cannot simply use $Blv$ with a single $v$.

Steps

1. Identify the geometry

   • Rod length: $L = 1\ \text{m}$
   • Rotation frequency: $f = 50\ \text{rev/s}$
   • Magnetic field: $B = 1\ \text{T}$, uniform and parallel to axis
   • One end at centre (hinged), other end slides on ring

2. Consider a small element

   Take a small segment of the rod at distance $r$ from the centre, of length $dr$.

   Its linear velocity: $v = \omega r$, where $\omega = 2\pi f = 100\pi\ \text{rad/s}$

3. EMF across the small element

   The motional emf across this segment:

4. Integrate over the entire rod The total emf between centre ($r=0$) and ring ($r=L$):

5. Substitute values

Final Answer

Why this works: Each radial segment of the rod cuts magnetic field lines at a speed proportional to its distance from the centre. The integration accounts for this varying speed, giving the correct net emf.

Common Mistakes in Motional EMF (Rotating Rod Problem)

Mistake 1: Using $v$ at the free end only

The error: Students often calculate emf as $E = B l v$, taking $v$ as the velocity of the free end ($v = \omega R$). This gives:

Why it's wrong: Different points on the rod move at different speeds. The free end moves fastest ($v = \omega R$), while the hinged end is stationary ($v = 0$). Using only the maximum velocity overestimates the emf.

How to avoid: Remember that for a rotating rod, each infinitesimal segment $dr$ at distance $r$ from the centre moves with velocity $v = \omega r$. The emf contributed by each segment is $dE = B \cdot v \cdot dr = B\omega r\, dr$. Integrate from $r = 0$ to $r = R$:

Mistake 2: Forgetting to convert frequency to angular velocity

The error: Students directly substitute $f = 50$ into formulas without converting to $\omega$.

Why it's wrong: The formula $v = \omega r$ requires angular velocity in rad/s, not rev/s.

How to avoid: Always convert: $\omega = 2\pi f = 2\pi \times 50 = 100\pi \ \text{rad/s}$. Then use:

Mistake 3: Confusing the rod with the ring

The error: Students try to find emf induced in the ring instead of the rod.

Why it's wrong: The ring is stationary and lies in a plane perpendicular to the magnetic field. A stationary loop in a uniform, constant field has zero induced emf (no change in flux).

How to avoid: Identify the moving conductor — it's the rod. The ring only serves as a sliding contact to complete the circuit. The emf is generated only in the rod as it cuts magnetic field lines.

Mistake 4: Forgetting the direction of emf (polarity)

The error: Students calculate magnitude but don't identify which end (centre or ring) is at higher potential.

Why it's wrong: Exam questions often ask for "emf between centre and ring" — this implies both magnitude and polarity.

How to avoid: Use Fleming's Right Hand Rule:

• Thumb: direction of motion of the rod (tangential)
• Forefinger: magnetic field direction (into/out of page)
• Middle finger: direction of induced current (and hence emf)

For this setup, the centre is at lower potential and the ring at higher potential (or vice versa depending on field direction — check carefully).

Mistake 5: Using wrong formula for average velocity

The error: Students take average velocity as $\frac{0 + \omega R}{2} = \frac{\omega R}{2}$ and then use $E = B l v_{avg}$.

Why it's wrong: While this gives the correct numerical answer ($\frac{1}{2}B\omega R^2$), it's conceptually sloppy and fails if the rod isn't straight or the field isn't uniform.

How to avoid: Always use the integration method — it's rigorous and works for all cases. The average velocity trick only works here by coincidence (linear variation of $v$ with $r$).

Quick Summary Checklist

Final correct answer: $\boxed{50\pi \ \text{V}}$ (approximately $157\ \text{V}$)