Q.A wheel with $10$ metallic spokes each $0.5\ \text{m}$ long is rotated with a speed of $120\ \text{rev/min}$ in a plane normal to the horizontal component of earth's magnetic field $H_E$ at a place. If $H_E = 0.4\ \text{G}$ at the place, what is the induced emf between the axle and the rim of the wheel? Note that $1\ \text{G} = 10^{-4}\ \text{T}$.

Concept: Motional Emf – each spoke acts as a rotating rod cutting magnetic field lines. The induced emf across a spoke is the same as that of a single rod rotating about one end.

Steps:

1. Angular speed:

   $\omega = 120\ \text{rev/min} = 120 \times \frac{2\pi}{60} = 4\pi\ \text{rad/s}$.

2. Magnetic field in SI:

   $H_E = 0.4\ \text{G} = 0.4 \times 10^{-4} = 4 \times 10^{-5}\ \text{T}$.

3. Motional emf for one spoke (length $L = 0.5\ \text{m}$):

   $\mathcal{E}_\text{spoke} = \frac{1}{2} B \omega L^2 = \frac{1}{2} \times (4 \times 10^{-5}) \times (4\pi) \times (0.5)^2$.

4. Compute: …

The key insight here is that each metallic spoke is like a rod rotating about one end in a uniform magnetic field. The axle and rim act as the two terminals, and all spokes are connected in parallel between them — so the net emf is just the emf of one spoke.

Why motional emf works here

When a conductor moves in a magnetic field, free electrons experience a magnetic force $q(\vec{v} \times \vec{B})$, which pushes them along the conductor until an electric field builds up to balance it. The resulting potential difference is the motional emf. For a rod rotating in a uniform field perpendicular to its plane, different parts of the rod move at different speeds — so we integrate the emf contribution along its length.

Step-by-step solution

1. Convert the rotation speed to angular velocity

The wheel rotates at $120\ \text{rev/min}$. One revolution is $2\pi$ radians, and one minute is $60$ seconds.

So $\omega = 4\pi \approx 12.57\ \text{rad/s}$.

2. Convert the magnetic field to tesla

Given $H_E = 0.4\ \text{G}$ and $1\ \text{G} = 10^{-4}\ \text{T}$:

3. Identify the length of a spoke

Each spoke is $L = 0.5\ \text{m}$ long, from axle (centre) to rim.

4. Apply the motional emf formula for a rotating rod

The emf induced in a single spoke is:

Substitute the values:

5. Simplify step by step

First, $(0.5)^2 = 0.25$.

Then $\frac{1}{2} \times 0.25 = 0.125$.

So: …

Method: Motional EMF in a Rotating Spoke (Faraday’s Law for a Conductor in a Magnetic Field)

This problem uses the concept that each metallic spoke acts like a rod rotating about one end in a uniform magnetic field. The induced emf is generated because the free electrons in the spoke experience a magnetic force due to their motion perpendicular to the field.

Steps

Step 1: Identify the geometry and direction

• The wheel rotates in a plane normal to the horizontal magnetic field $H_E$.
• Each spoke is a conductor of length $L = 0.5\ \text{m}$, rotating about the axle (one end fixed).
• The magnetic field is perpendicular to the plane of rotation → the velocity of each point on the spoke is perpendicular to both the spoke and the field.

Step 2: Convert given data to SI units

• Magnetic field:

• Angular speed:

Step 3: Use the formula for motional emf in a rotating rod

For a rod of length $L$ rotating with angular speed $\omega$ about one end in a uniform perpendicular magnetic field $B$, the induced emf between the centre (axle) and the rim is:

Why?

• The linear velocity of a point at distance $r$ from the axle is $v = \omega r$.
• The motional emf across a small element $dr$ is $d\mathcal{E} = B v\, dr = B \omega r\, dr$.
• Integrate from $r = 0$ to $r = L$: …

Here are the most common mistakes students make on this classic Motional EMF problem, along with how to avoid each.

1. Mistake: Treating the spokes as separate generators in parallel

The error:

Students often calculate the EMF for one spoke, then multiply by 10 (the number of spokes), thinking each adds to the total voltage.

Why it’s wrong:

All spokes are connected to the same axle (one terminal) and the same rim (the other terminal). They are electrically in parallel, not in series. The induced EMF between axle and rim is the same as that of a single spoke.

How to avoid:

• Visualise the circuit: axle → spoke → rim → back through another spoke → axle.
• Parallel branches have the same potential difference.
• Key result: Total EMF = EMF of one spoke only.

2. Mistake: Using the wrong formula for motional EMF in a rotating rod

The error:

Using $\mathcal{E} = B l v$, where $v$ is the linear speed of the tip, or using $v$ at the midpoint incorrectly.

Why it’s wrong:

For a rod rotating about one end, different points move at different speeds ($v = \omega r$). The correct formula integrates the EMF along the length:

How to avoid:

• Derive quickly: For a small element $dr$ at distance $r$ from axle, $d\mathcal{E} = B (\omega r) dr$. Integrate from $0$ to $L$:

• Memorise this form for a rod rotating about one end in a uniform perpendicular field.

3. Mistake: Forgetting unit conversion for magnetic field

The error:

Plugging $H_E = 0.4\ \text{G}$ directly into the formula without converting to tesla.

Why it’s wrong:

The formula $\mathcal{E} = \frac{1}{2} B \omega L^2$ requires $B$ in tesla (T).

Given $1\ \text{G} = 10^{-4}\ \text{T}$,

How to avoid:

• Always write the conversion step explicitly.
• Check units: if $B$ is in gauss, convert before substituting.

4. Mistake: Using wrong angular velocity units

The error:

Using $120\ \text{rev/min}$ directly as $\omega$ in rad/s without conversion.

Why it’s wrong:

$\omega$ must be in radians per second for the formula to work.

Correct conversion:

How to avoid:

• Write the conversion chain every time. …