Q.Kamla peddles a stationary bicycle. The pedals of the bicycle are attached to a $100$ turn coil of area $0.10\ \text{m}^2$. The coil rotates at half a revolution per second and it is placed in a uniform magnetic field of $0.01\ \text{T}$ perpendicular to the axis of rotation of the coil. What is the maximum voltage generated in the coil?

Concept: Motional EMF — specifically, the sinusoidal voltage induced in a rotating coil within a uniform magnetic field (an AC generator).

Reasoning:

1. The instantaneous EMF in a rotating coil is given by

   $\mathcal{E} = N B A \omega \sin(\omega t)$, where $N$ is the number of turns, $B$ the magnetic field, $A$ the area, and $\omega$ the angular speed.

2. The maximum voltage occurs when $\sin(\omega t) = 1$, so

   $\mathcal{E}_{\text{max}} = N B A \omega$.

3. Given: $N = 100$, $B = 0.01\ \text{T}$, $A = 0.10\ \text{m}^2$, and frequency $f = 0.5\ \text{Hz}$. …

This is a classic AC generator setup. When a coil rotates in a uniform magnetic field, the magnetic flux through it changes sinusoidally, inducing an alternating EMF. The key is to connect the physical rotation to Faraday’s law.

The coil has $N$ turns, area $A$, rotates at frequency $f$ in field $B$. The flux at any instant is $\Phi = NBA\cos\theta$, where $\theta = \omega t$ (if the coil starts perpendicular to the field). Faraday’s law gives $\mathcal{E} = -d\Phi/dt$, which yields a sine wave. The maximum EMF occurs when the rate of change of flux is greatest — that’s when the coil is parallel to the field ($\theta = 90^\circ$ or $270^\circ$).

The formula $\mathcal{E}_0 = NBA\omega$ is the standard result. Let’s apply it step by step.

1. Identify the given data

   • Number of turns: $N = 100$
   • Area of coil: $A = 0.10\ \text{m}^2$
   • Magnetic field: $B = 0.01\ \text{T}$
   • Frequency of rotation: $f = 0.5\ \text{rev/s}$ (half a revolution per second)

2. Find the angular frequency $\omega$

   One revolution is $2\pi$ radians. So $\omega = 2\pi f = 2\pi \times 0.5 = \pi\ \text{rad/s}$.

   Tip

   Always convert “revolutions per second” to rad/s by multiplying by $2\pi$. Don’t forget — many students use $f$ directly in the formula and get the wrong unit.

3. Apply the maximum EMF formula

   For a rotating coil in a uniform field, the induced EMF is $\mathcal{E} = NBA\omega \sin(\omega t)$. The maximum value is $\mathcal{E}_0 = NBA\omega$.

   $\mathcal{E}_0 = N B A \omega$ …

Method: Faraday's Law of Electromagnetic Induction (for a Rotating Coil in a Uniform Magnetic Field)

This is a classic AC generator problem where a coil rotates in a uniform magnetic field, producing a sinusoidal emf.

Steps:

Step 1: Identify the key formula

For a coil of $N$ turns, area $A$, rotating with angular frequency $\omega$ in a uniform magnetic field $B$, the instantaneous induced emf is:

The maximum emf occurs when $\sin(\omega t) = 1$:

Step 2: Extract given data

• Number of turns, $N = 100$
• Area of coil, $A = 0.10\ \text{m}^2$
• Magnetic field, $B = 0.01\ \text{T}$
• Frequency of rotation, $f = 0.5\ \text{rev/s}$ (half a revolution per second)

Step 3: Convert frequency to angular frequency

Step 4: Substitute into the maximum emf formula

$\mathcal{E}_{\text{max}} = 100 \times 0.01 \times 0.10 \times \pi$ …

Common Mistakes in Motional EMF Problems (Kamla's Bicycle)

Mistake 1: Confusing Frequency ($f$) with Angular Frequency ($\omega$)

The error: Students often plug $f = 0.5\ \text{Hz}$ directly into the formula without converting to $\omega$.

Why it's wrong: The induced EMF formula uses angular frequency:

where $\omega = 2\pi f$, not $f$ itself.

How to avoid: Always convert: $\omega = 2\pi f = 2\pi(0.5) = \pi\ \text{rad/s}$

Mistake 2: Forgetting the Number of Turns ($N$)

The error: Using $N=1$ instead of $N=100$.

Why it's wrong: Each turn contributes independently to the induced EMF. The total EMF is $N$ times the EMF of a single turn.

How to avoid: Circle the number of turns in the problem statement. Write the formula as $\mathcal{E}_{\text{max}} = N B A \omega$ before substituting.

Mistake 3: Misinterpreting "Half a Revolution per Second"

The error: Taking $f = 0.5$ as the angular speed directly, or thinking "half revolution" means $\theta = \pi/2$.

Why it's wrong: "Half a revolution per second" means frequency = 0.5 Hz (0.5 complete cycles per second), not an angle.

How to avoid: Remember: "revolutions per second" = frequency ($f$). Convert to $\omega$ using $\omega = 2\pi f$.

Mistake 4: Using Peak-to-Peak or RMS Instead of Maximum

The error: Reporting $\mathcal{E}_{\text{rms}}$ or $\mathcal{E}_{\text{avg}}$ instead of $\mathcal{E}_{\text{max}}$.

Why it's wrong: The question explicitly asks for maximum voltage. For sinusoidal EMF:

• $\mathcal{E}_{\text{max}} = NBA\omega$
• $\mathcal{E}_{\text{rms}} = \frac{\mathcal{E}_{\text{max}}}{\sqrt{2}}$

How to avoid: Read carefully — "maximum" means the peak value. Use $\sin(\omega t) = 1$ to get $\mathcal{E}_{\text{max}}$.

Mistake 5: Incorrect Area or Magnetic Field Units …