Electromagnetic Induction – Study Notes
Program: NSO (Science Olympiad)
Subject: Physics
Topic: Electromagnetic Induction
These notes are written for quick reference and exam‑style problem solving. They are organized into a short explanation, key facts, a worked example, and a revision summary.
1. Simple Explanation
When a conductor moves inside a magnetic field, or when the magnetic field that links a circuit changes, an electromotive force (EMF) is generated.
Faraday’s Law tells us the magnitude of that EMF, while Lenz’s Law explains its direction: the induced current always opposes the change that produced it.
2. Key Points
| Concept | Formula / Expression | Units | Typical Notes |
|---|---|---|---|
| Magnetic Flux (Φ) | ( \Phi = \int \mathbf{B}\cdot d\mathbf{A} ) | webers (Wb) | For a uniform field over a flat area: ( \Phi = B,A\cos\theta ) |
| Faraday’s Law (Induced EMF) | ( \mathcal{E} = -N,\frac{d\Phi}{dt} ) | volts (V) | (N) = number of turns, negative sign = Lenz’s law |
| Moving Conductor in Uniform Field | ( \mathcal{E} = B,\ell,v ) | V | ( \ell ) = length of conductor, (v) = velocity perpendicular to (B) |
| Inductance (Self‑) | ( \mathcal{E} = -L,\frac{dI}{dt} ) | Henry (H) | ( L = \frac{N\Phi}{I} ) |
| Mutual Inductance | ( \mathcal{E}_2 = -M,\frac{dI_1}{dt} ) | H | ( M ) = coupling between two coils |
| Lenz’s Law | Direction of induced current ⇔ opposes change in flux | — | Expressed graphically: “negative sign in Faraday’s law” |
| Units | B = Tesla (T), Φ = Weber (Wb), I = Ampere (A), L = Henry (H) | — | 1 T = 1 Wb/m² |
| Maxwell’s Equation (Faraday) | ( \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} ) | — | Gives the curl of the electric field |
Remember:
- Flux change can come from moving the conductor, changing B, or changing the area/angle.
- Sign convention: The negative sign in Faraday’s law is crucial for predicting direction.
- Inductors store magnetic energy: ( U = \frac{1}{2}LI^2 ).
3. Worked Example (Generator Style)
Problem
A coil of 50 turns is rotated in a uniform magnetic field of 0.4 T.
The coil has a radius of 0.05 m and rotates at 720 rpm (12 rev/s).
Assume the coil is a perfect circle, and the axis of rotation is perpendicular to the field.
What is the maximum induced EMF across the coil?
Step‑by‑Step Solution
-
Find the area of the coil
[ A = \pi r^2 = \pi(0.05,\text{m})^2 \approx 7.85\times10^{-3},\text{m}^2 ] -
Flux through one turn (when coil is perpendicular)
[ \Phi_{\max} = B,A = 0.4,\text{T}\times 7.85\times10^{-3},\text{m}^2 = 3.14\times10^{-3},\text{Wb} ] -
Flux as a function of time
When the coil rotates, the flux varies sinusoidally: [ \Phi(t) = \Phi_{\max}\cos(\omega t) ] where (\omega = 2\pi f) and (f = 12,\text{Hz}). -
Differentiate to get EMF
[ \mathcal{E}(t) = -N,\frac{d\Phi}{dt} = -N,(-\Phi_{\max}\omega\sin\omega t) = N\Phi_{\max}\omega\sin\omega t ] -
Maximum EMF (when (\sin\omega t = 1))
[ \mathcal{E}{\max} = N,\Phi{\max},\omega ] [ \