Movie Hook: The Arc Reactor in Iron Man
Remember Tony Stark’s arc reactor? That glowing device in his chest isn’t just science fiction - it’s based on Faraday’s laws of electromagnetic induction! When magnetic flux changes through a coil, it generates electricity. The arc reactor uses rotating magnetic fields to induce massive currents, powering the Iron Man suit. Every time you charge your phone wirelessly or use a power generator, you’re using Faraday’s discovery from 1831.
The Discovery That Changed the World
In 1831, Michael Faraday discovered that electricity and magnetism are two sides of the same coin. If Oersted showed that current creates magnetism (1820), Faraday proved the reverse: changing magnetism creates current.
Faraday’s Experimental Setup
Faraday wound two separate coils around an iron ring. When he connected one coil to a battery, the other coil (not connected to anything!) showed a momentary deflection in a galvanometer. The key insight: the deflection occurred only when the current was changing - when switching on or off.
Faraday’s First Law
Whenever the magnetic flux linked with a coil changes, an electromotive force (EMF) is induced in the coil.
The induced EMF exists only as long as the flux is changing. No change in flux = no induced EMF.
Ways to Change Magnetic Flux
- Change the magnetic field strength (B): Move a magnet toward/away from a coil
- Change the area (A): Expand/compress a loop in a magnetic field
- Change the orientation (θ): Rotate a coil in a magnetic field
All three methods appear in JEE problems!
Faraday’s Second Law
The magnitude of induced EMF is directly proportional to the rate of change of magnetic flux.
Mathematically:
Where:
- $\mathcal{E}$ = Induced EMF (volts)
- $N$ = Number of turns in coil
- $\Phi_B$ = Magnetic flux = $\vec{B} \cdot \vec{A} = BA\cos\theta$
- The negative sign comes from Lenz’s law
Understanding the Formula
Flux linkage: For N turns, total flux linkage = $N\Phi_B$
The faster you change the flux, the greater the induced EMF. This is why:
- Rapidly moving a magnet induces more EMF than slowly moving it
- More turns give more EMF (each turn contributes)
- Generators spin at high speeds to maximize $\frac{d\Phi}{dt}$
Memory Trick: “FRAC”
Faraday’s law: Rate of Area-field Change
The induced EMF depends on how fast you change the “area-field” product ($BA$).
Magnetic Flux: The Core Concept
Think of magnetic flux as “how many magnetic field lines pass through a surface.”
For uniform field and flat surface:
- Maximum flux: θ = 0° (surface perpendicular to field)
- Zero flux: θ = 90° (surface parallel to field)
Units:
- Flux: Weber (Wb) = T·m² = V·s
- EMF: Volt (V) = Wb/s
Types of Problems in JEE
Type 1: Magnet Moving Through Coil
Setup: Bar magnet moves through a coil with N turns, area A.
Solution Strategy:
- Identify how B changes with position/time
- Calculate $\Phi_B = BA\cos\theta$ at different positions
- Find $\mathcal{E} = -N\frac{d\Phi_B}{dt}$
Type 2: Rotating Coil in Magnetic Field
Setup: Rectangular coil rotates with angular velocity ω in uniform field B.
This is the AC generator principle (see AC circuits)!
Type 3: Changing Area
Setup: Conducting rod slides on rails in magnetic field.
Key: Area increases as $A = \ell x$ where x changes with time.
$$\mathcal{E} = -\frac{d\Phi_B}{dt} = -B\frac{dA}{dt} = -B\ell\frac{dx}{dt} = -B\ell v$$Common Mistakes Students Make
❌ Mistake 1: Thinking static magnetic fields induce EMF ✅ Reality: Only changing flux induces EMF
❌ Mistake 2: Forgetting the cosθ factor in flux ✅ Reality: $\Phi_B = BA\cos\theta$ - angle matters!
❌ Mistake 3: Using the wrong angle ✅ Reality: θ is angle between $\vec{B}$ and area vector $\vec{A}$ (perpendicular to surface)
❌ Mistake 4: Confusing flux with flux density ✅ Reality:
- Flux density = B (Tesla)
- Flux = Φ = BA (Weber)
❌ Mistake 5: Ignoring the number of turns ✅ Reality: EMF is proportional to N
Interactive Exploration
Try changing:
- Speed of magnet motion
- Number of coil turns
- Strength of magnet
Observe how induced EMF varies!
Practice Problems
Level 1: JEE Main (Basics)
Problem 1.1: A circular coil of 100 turns and radius 5 cm is placed perpendicular to a magnetic field of 0.2 T. If the field is reduced to zero in 0.1 s, find the induced EMF.
Solution
Given: N = 100, r = 5 cm = 0.05 m, B₁ = 0.2 T, B₂ = 0, Δt = 0.1 s
Area: $A = \pi r^2 = \pi(0.05)^2 = 7.85 \times 10^{-3}$ m²
Initial flux: $\Phi_1 = B_1A = 0.2 \times 7.85 \times 10^{-3} = 1.57 \times 10^{-3}$ Wb
Final flux: $\Phi_2 = 0$
Change in flux: $\Delta\Phi = \Phi_2 - \Phi_1 = -1.57 \times 10^{-3}$ Wb
Induced EMF:
$$|\mathcal{E}| = N\frac{|\Delta\Phi|}{\Delta t} = 100 \times \frac{1.57 \times 10^{-3}}{0.1} = 1.57 \text{ V}$$Answer: 1.57 V
Problem 1.2: A coil of 200 turns and area 0.01 m² is rotating at 50 rev/s in a magnetic field of 0.5 T. Find the maximum induced EMF.
Solution
Given: N = 200, A = 0.01 m², f = 50 rev/s, B = 0.5 T
Angular velocity: $\omega = 2\pi f = 2\pi \times 50 = 100\pi$ rad/s
Maximum EMF in rotating coil:
$$\mathcal{E}_{max} = NBA\omega = 200 \times 0.5 \times 0.01 \times 100\pi$$ $$\mathcal{E}_{max} = 100\pi = 314.16 \text{ V}$$Answer: 314 V
Level 2: JEE Advanced (Application)
Problem 2.1: A square loop of side 10 cm is placed with its plane perpendicular to a magnetic field. The field varies as $B = 0.2t^2$ (in Tesla, t in seconds). Calculate the induced EMF at t = 5 s.
Solution
Given: Side a = 10 cm = 0.1 m, $B = 0.2t^2$ T
Area: $A = a^2 = (0.1)^2 = 0.01$ m² (constant)
Flux: $\Phi_B = BA = 0.2t^2 \times 0.01 = 2 \times 10^{-3}t^2$ Wb
Induced EMF:
$$\mathcal{E} = -\frac{d\Phi_B}{dt} = -\frac{d}{dt}(2 \times 10^{-3}t^2) = -4 \times 10^{-3}t$$At t = 5 s:
$$|\mathcal{E}| = 4 \times 10^{-3} \times 5 = 0.02 \text{ V} = 20 \text{ mV}$$Answer: 20 mV
Problem 2.2: A conducting rod of length 20 cm moves with velocity 5 m/s perpendicular to a magnetic field of 0.8 T. The rod, field, and velocity are mutually perpendicular. Find the induced EMF.
Solution
This is motional EMF (will be covered in detail in motional-emf.md)
Given: ℓ = 20 cm = 0.2 m, v = 5 m/s, B = 0.8 T
For a straight conductor moving perpendicular to B:
$$\mathcal{E} = B\ell v = 0.8 \times 0.2 \times 5 = 0.8 \text{ V}$$Answer: 0.8 V
Level 3: JEE Advanced (Challenging)
Problem 3.1: A triangular loop ABC is placed in a non-uniform magnetic field given by $\vec{B} = B_0x\hat{k}$ where x is horizontal distance. The triangle has vertices at A(0,0), B(a,0), C(a/2, h). Find the magnetic flux through the triangle.
Solution
Since B varies with x, we must integrate:
$$\Phi_B = \int \vec{B} \cdot d\vec{A}$$For a vertical strip at distance x of width dx:
- Height of strip at position x: $y(x) = \begin{cases} 2hx/a & 0 \leq x \leq a/2 \\ 2h(a-x)/a & a/2 \leq x \leq a \end{cases}$
Strip area: $dA = y(x)dx$
Flux through strip: $d\Phi = B_0x \cdot y(x)dx$
Due to symmetry, we can calculate for first half and double:
$$\Phi_B = 2\int_0^{a/2} B_0x \cdot \frac{2hx}{a}dx = \frac{4B_0h}{a}\int_0^{a/2} x^2dx$$ $$\Phi_B = \frac{4B_0h}{a} \cdot \frac{x^3}{3}\Big|_0^{a/2} = \frac{4B_0h}{a} \cdot \frac{a^3}{24} = \frac{B_0ha^2}{6}$$Answer: $\Phi_B = \frac{B_0ha^2}{6}$
Problem 3.2: A circular loop of radius R is placed in a time-varying magnetic field $B = B_0(1 + \alpha t)$ perpendicular to its plane. A resistance r is connected across a diameter. Find the current through the resistor.
Solution
Given: Radius R, $B = B_0(1 + \alpha t)$, resistance r across diameter
Area of circle: $A = \pi R^2$
Flux: $\Phi_B = BA = B_0(1 + \alpha t)\pi R^2$
Induced EMF:
$$\mathcal{E} = -\frac{d\Phi_B}{dt} = -B_0\alpha\pi R^2$$Key insight: The resistor across the diameter divides the loop into two halves. Each half acts as a battery with EMF $\mathcal{E}/2$ in series with internal resistance.
However, for this configuration, the two halves are in parallel with opposite polarity. The net effect is that current flows through the diameter.
Each semicircular arc has resistance $R_{arc} = \rho \frac{\pi R}{A_{wire}}$
Using symmetry and circuit analysis:
$$I = \frac{|\mathcal{E}|}{2r} = \frac{B_0\alpha\pi R^2}{2r}$$Answer: $I = \frac{B_0\alpha\pi R^2}{2r}$
(Note: Complete analysis requires knowing the wire resistance, which affects the result. In ideal case with zero wire resistance, answer is as above.)
Connection to Other Chapters
- Magnetism: Magnetic field B is the source that creates flux
- Current Electricity: Induced EMF drives current through Ohm’s law
- Lenz’s Law: Explains the negative sign in Faraday’s law
- Motional EMF: Special case where conductor moves in field
- Conservation of Energy: Energy conversion in induction
Real-World Applications
- Electric Generators: Convert mechanical energy to electrical (power plants)
- Wireless Charging: Your phone uses changing magnetic fields
- Transformers: Change AC voltage levels (covered in transformers.md)
- Induction Cooktops: Heat metal pans using induced currents
- Metal Detectors: Detect induced currents in metal objects
- Electromagnetic Braking: Used in trains and roller coasters
Key Takeaways
✓ EMF is induced only when magnetic flux changes ✓ Induced EMF = $-N\frac{d\Phi_B}{dt}$ (fundamental formula) ✓ Three ways to change flux: change B, A, or θ ✓ More turns → more EMF (directly proportional to N) ✓ Faster change → more EMF (rate matters) ✓ The negative sign indicates direction (Lenz’s law)
Quick Revision Formula Sheet
| Quantity | Formula | Units |
|---|---|---|
| Magnetic Flux | $\Phi_B = BA\cos\theta$ | Wb (Weber) |
| Induced EMF | $\mathcal{E} = -N\frac{d\Phi_B}{dt}$ | V (Volt) |
| Rotating Coil | $\mathcal{E} = NBA\omega\sin(\omega t)$ | V |
| Max EMF (AC) | $\mathcal{E}_{max} = NBA\omega$ | V |
| Motional EMF | $\mathcal{E} = B\ell v$ | V |
Next Topic: Lenz’s Law - Understanding the direction of induced current