Electromagnetic Induction

Faraday's law, Lenz's law, motional EMF, generators, transformers

⚡ Electromagnetic Induction

Magnetic Flux

Magnetic flux $\Phi_B$ through a surface:

$\Phi_B = \vec{B} \cdot \vec{A} = BA \cos\theta$

where:

B = magnetic field strength (T)
A = area (m²)
$\theta$ = angle between B and normal to surface

Unit: Weber (Wb) = T·m²

Maximum flux: B perpendicular to surface ( $\theta = 0°$ , $\Phi = BA$ ) Zero flux: B parallel to surface ( $\theta = 90°$ )

Faraday's Law

Changing magnetic flux induces EMF (voltage)!

$\mathcal{E} = -N\frac{\Delta \Phi_B}{\Delta t}$

where:

$\mathcal{E}$ = induced EMF (V)
N = number of turns in coil
$\Delta \Phi_B$ = change in magnetic flux

Three ways to change flux:

Change B (vary field strength)
Change A (vary area)
Change θ (rotate coil)

💡 This is how generators work! Motion → changing flux → induced voltage

Lenz's Law

The direction of induced current opposes the change that caused it.

Negative sign in Faraday's Law represents this!

To find direction:

Determine if flux is increasing or decreasing
Induced current creates B field to oppose this change
Use right-hand rule to find current direction

Increasing flux: Induced B field opposes (points opposite) Decreasing flux: Induced B field reinforces (points same way)

💡 Conservation of energy! If induced current helped the change, you'd get free energy (perpetual motion).

Motional EMF

Conductor moving through B field:

$\mathcal{E} = BLv$

where:

B = field strength
L = length of conductor
v = velocity perpendicular to B

Physical picture:

Moving conductor → charges inside move
Magnetic force on charges: F = qvB
Charges separate → voltage (EMF)!

Direction: Use right-hand rule for F = qv × B

Applications: Electric Generator

Rotate coil in magnetic field
Flux changes: $\Phi(t) = BA \cos(\omega t)$
Induced EMF: $\mathcal{E}(t) = NBA\omega \sin(\omega t)$

AC generator: Produces alternating current (sinusoidal)

DC generator: Uses commutator to rectify current (one direction)

Maximum EMF: $\mathcal{E}_{max} = NBA\omega$

Eddy Currents

Induced currents in solid conductors:

Swirling currents (like water eddies)
Dissipate energy as heat
Create magnetic braking force

Applications:

Braking systems (trains)
Metal detectors
Induction cooktops

Reduce eddy currents: Use laminated (layered) cores

Transformers

Step-up/step-down voltage using induction!

$\frac{V_s}{V_p} = \frac{N_s}{N_p}$

where:

$V_p$ , $N_p$ = primary voltage, turns
$V_s$ , $N_s$ = secondary voltage, turns

Power conservation (ideal transformer): $P_p = P_s$ $V_p I_p = V_s I_s$

So: $\frac{I_s}{I_p} = \frac{V_p}{V_s} = \frac{N_p}{N_s}$

Step-up (N_s > N_p): Voltage increases, current decreases Step-down (N_s < N_p): Voltage decreases, current increases

💡 Only works with AC! Need changing flux.

Self-Inductance

Changing current in coil induces EMF in same coil!

$\mathcal{E} = -L\frac{\Delta I}{\Delta t}$

where L is inductance (unit: Henry, H = Wb/A = V·s/A)

For solenoid: $L = \mu_0 n^2 A \ell = \mu_0 \frac{N^2 A}{\ell}$

Opposes change in current (like inertia for current!)

Energy in Magnetic Field

Energy stored in inductor: $U_B = \frac{1}{2}LI^2$

Energy density in B field: $u_B = \frac{B^2}{2\mu_0}$

RL Circuits

Inductor in circuit with resistor:

Growth of current (switch closed): $I(t) = I_{max}(1 - e^{-t/\tau})$

where $\tau = L/R$ is time constant

Decay of current (switch opened): $I(t) = I_0 e^{-t/\tau}$

After time τ: Current reaches 63% of maximum (or decays to 37%)

Maxwell's Addition to Ampère's Law

Changing electric field creates magnetic field!

Just as changing B creates E (Faraday), changing E creates B.

This led to prediction of electromagnetic waves!

$c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} = 3.0 \times 10^8 \text{ m/s}$

Speed of light emerges from electric and magnetic constants!

Problem-Solving Strategy

Find initial and final flux: $\Phi = BA \cos\theta$
Calculate change: $\Delta \Phi$
Apply Faraday's Law: $\mathcal{E} = -N\Delta\Phi/\Delta t$
Use Lenz's Law for direction
For motional EMF: $\mathcal{E} = BLv$

Common Mistakes

❌ Forgetting cos θ in flux calculation ❌ Wrong sign/direction from Lenz's Law ❌ Using Faraday's Law when flux is constant (ΔΦ = 0 → ε = 0!) ❌ Confusing transformers (works with AC only!) ❌ Thinking V increases → I increases in transformer (opposite!) ❌ Not using perpendicular component of velocity

📚 Practice Problems

1Problem 1easy

❓ Question:

A circular coil with 50 turns and radius 10 cm is perpendicular to a 0.80 T magnetic field. The field decreases to zero in 0.20 s. What is the induced EMF?

💡 Show Solution

Given:

Number of turns: $N = 50$
Radius: $r = 10$ cm $= 0.10$ m
Initial field: $B_i = 0.80$ T
Final field: $B_f = 0$ T
Time: $\Delta t = 0.20$ s
Perpendicular: $\theta = 0°$

Solution:

Step 1: Find area. $A = \pi r^2 = \pi (0.10)^2 = 0.0314 \text{ m}^2$

Step 2: Find change in flux. $\Phi_i = B_i A \cos\theta = (0.80)(0.0314)(1) = 0.0251 \text{ Wb}$ $\Phi_f = 0$ $\Delta \Phi = \Phi_f - \Phi_i = -0.0251 \text{ Wb}$

Step 3: Apply Faraday's Law. $|\mathcal{E}| = N\frac{|\Delta \Phi|}{\Delta t} = (50)\frac{0.0251}{0.20}$ $|\mathcal{E}| = 6.3 \text{ V}$

Answer: Induced EMF = 6.3 V

Direction: By Lenz's Law, induced current creates B field in same direction as original (to oppose the decrease).

2Problem 2easy

❓ Question:

A circular coil with 50 turns and radius 10 cm is perpendicular to a 0.80 T magnetic field. The field decreases to zero in 0.20 s. What is the induced EMF?

💡 Show Solution

Given:

Number of turns: $N = 50$
Radius: $r = 10$ cm $= 0.10$ m
Initial field: $B_i = 0.80$ T
Final field: $B_f = 0$ T
Time: $\Delta t = 0.20$ s
Perpendicular: $\theta = 0°$

Solution:

Step 1: Find area. $A = \pi r^2 = \pi (0.10)^2 = 0.0314 \text{ m}^2$

Step 2: Find change in flux. $\Phi_i = B_i A \cos\theta = (0.80)(0.0314)(1) = 0.0251 \text{ Wb}$ $\Phi_f = 0$ $\Delta \Phi = \Phi_f - \Phi_i = -0.0251 \text{ Wb}$

Step 3: Apply Faraday's Law. $|\mathcal{E}| = N\frac{|\Delta \Phi|}{\Delta t} = (50)\frac{0.0251}{0.20}$ $|\mathcal{E}| = 6.3 \text{ V}$

Answer: Induced EMF = 6.3 V

Direction: By Lenz's Law, induced current creates B field in same direction as original (to oppose the decrease).

3Problem 3medium

❓ Question:

A metal rod of length 0.50 m moves at 4.0 m/s perpendicular to a 0.30 T magnetic field. What is the motional EMF?

💡 Show Solution

Given:

Length: $L = 0.50$ m
Velocity: $v = 4.0$ m/s
B field: $B = 0.30$ T
Perpendicular motion

Solution:

Motional EMF: $\mathcal{E} = BLv = (0.30)(0.50)(4.0) = 0.60 \text{ V}$

Physical picture:

Rod moves through B field
Free electrons in metal experience F = qvB
Electrons accumulate at one end
Creates potential difference (EMF)!

Answer: Motional EMF = 0.60 V

Direction: Use right-hand rule for F = qv × B (remembering q is negative for electrons).

4Problem 4medium

❓ Question:

A metal rod of length 0.50 m moves at 4.0 m/s perpendicular to a 0.30 T magnetic field. What is the motional EMF?

💡 Show Solution

Given:

Length: $L = 0.50$ m
Velocity: $v = 4.0$ m/s
B field: $B = 0.30$ T
Perpendicular motion

Solution:

Motional EMF: $\mathcal{E} = BLv = (0.30)(0.50)(4.0) = 0.60 \text{ V}$

Physical picture:

Rod moves through B field
Free electrons in metal experience F = qvB
Electrons accumulate at one end
Creates potential difference (EMF)!

Answer: Motional EMF = 0.60 V

Direction: Use right-hand rule for F = qv × B (remembering q is negative for electrons).

5Problem 5hard

❓ Question:

An ideal transformer steps 120 V AC down to 12 V. The primary has 100 turns. (a) How many turns in secondary? (b) If secondary supplies 5.0 A, what is primary current?

💡 Show Solution

Given:

Primary voltage: $V_p = 120$ V
Secondary voltage: $V_s = 12$ V
Primary turns: $N_p = 100$
Secondary current: $I_s = 5.0$ A

Part (a): Secondary turns

$\frac{V_s}{V_p} = \frac{N_s}{N_p}$ $N_s = N_p \frac{V_s}{V_p} = (100)\frac{12}{120} = 10 \text{ turns}$

Part (b): Primary current

Power conservation: $V_p I_p = V_s I_s$ $I_p = I_s \frac{V_s}{V_p} = (5.0)\frac{12}{120} = 0.50 \text{ A}$

Check: Voltage down by factor of 10, so current up by factor of 10! ✓

Answer:

(a) N_s = 10 turns (step-down transformer)
(b) I_p = 0.50 A

Note: Lower voltage → higher current for same power.

6Problem 6hard

❓ Question:

An ideal transformer steps 120 V AC down to 12 V. The primary has 100 turns. (a) How many turns in secondary? (b) If secondary supplies 5.0 A, what is primary current?

💡 Show Solution

Given:

Primary voltage: $V_p = 120$ V
Secondary voltage: $V_s = 12$ V
Primary turns: $N_p = 100$
Secondary current: $I_s = 5.0$ A

Part (a): Secondary turns

$\frac{V_s}{V_p} = \frac{N_s}{N_p}$ $N_s = N_p \frac{V_s}{V_p} = (100)\frac{12}{120} = 10 \text{ turns}$

Part (b): Primary current

Power conservation: $V_p I_p = V_s I_s$ $I_p = I_s \frac{V_s}{V_p} = (5.0)\frac{12}{120} = 0.50 \text{ A}$

Check: Voltage down by factor of 10, so current up by factor of 10! ✓

Answer:

(a) N_s = 10 turns (step-down transformer)
(b) I_p = 0.50 A

Note: Lower voltage → higher current for same power.

🎴

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