Faraday's Law and Lenz's Law

Electromagnetic induction and induced EMF

Faraday's Law and Lenz's Law

Magnetic Flux

$\Phi_B = \int \vec{B} \cdot d\vec{A}$

For uniform field and flat surface: $\Phi_B = BA\cos\theta$

Units: 1 weber (Wb) = 1 T·m²

Faraday's Law

Induced EMF: $\mathcal{E} = -\frac{d\Phi_B}{dt}$

Negative sign is Lenz's law: induced current opposes change in flux.

For N-turn coil: $\mathcal{E} = -N\frac{d\Phi_B}{dt}$

Ways to Change Flux

Change B: $\Phi_B = BA\cos\theta$ , vary $B$
Change A: Move wire to change loop area
Change θ: Rotate loop in field

Motional EMF

Rod of length $L$ moving with velocity $v$ perpendicular to field $B$ :

Magnetic force on charges: $F = qvB$

EMF: $\mathcal{E} = BLv$

From Faraday's law: $\mathcal{E} = -\frac{d(BA)}{dt} = -B\frac{dA}{dt} = -BL\frac{dx}{dt} = -BLv$

(Magnitude $BLv$ ; sign from Lenz's law)

Induced Electric Field

Changing magnetic flux creates electric field:

$\oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi_B}{dt}$

This $\vec{E}$ is non-conservative! (Circulation not zero)

For cylindrical symmetry: $E(2\pi r) = -\frac{d\Phi_B}{dt}$

Eddy Currents

Induced currents in bulk conductor:

Flow in loops (eddies)
Dissipate energy as heat
Create magnetic braking

Applications:

Metal detectors
Magnetic braking
Induction heating

Generators

Rotating coil in magnetic field:

$\Phi_B = BA\cos(\omega t)$

$\mathcal{E} = -\frac{d\Phi_B}{dt} = BA\omega\sin(\omega t)$

$\mathcal{E} = \mathcal{E}_0\sin(\omega t)$

where $\mathcal{E}_0 = NBA\omega$ (N turns, area A).

Betatron

Accelerates electrons in circular path:

Condition for stable orbit: $B_{orbit} = \frac{1}{2}B_{avg}$

where $B_{avg}$ is average field inside orbit.

Lenz's Law

Induced current creates magnetic field that opposes the change in flux.

Flux increasing: induced $\vec{B}$ opposes it
Flux decreasing: induced $\vec{B}$ tries to maintain it

Energy conservation: Work required to change flux against induced current.

📚 Practice Problems

1Problem 1medium

❓ Question:

A circular loop of radius r = 0.08 m has resistance R = 2.0 Ω. The loop is perpendicular to a uniform magnetic field that increases from B₀ = 0.2 T to B₁ = 0.8 T in time Δt = 0.5 s. Find: (a) the induced EMF, (b) the induced current (magnitude and direction), and (c) the total charge that flows through the loop.

💡 Show Solution

Given:

r = 0.08 m
R = 2.0 Ω
B₀ = 0.2 T, B₁ = 0.8 T
Δt = 0.5 s
Loop perpendicular to field

(a) Induced EMF:

Area: $A = \pi r^2 = \pi(0.08)^2 = 0.0201$ m²

Change in flux: $\Delta \Phi_B = A \Delta B = A(B_1 - B_0)$ $\Delta \Phi_B = (0.0201)(0.8 - 0.2) = 0.0121 \text{ Wb}$

Faraday's law: $|\mathcal{E}| = \left|\frac{\Delta \Phi_B}{\Delta t}\right| = \frac{0.0121}{0.5}$

$|\mathcal{E}| = \boxed{0.0241 \text{ V} = 24.1 \text{ mV}}$

(b) Induced current:

$I = \frac{|\mathcal{E}|}{R} = \frac{0.0241}{2.0}$

$I = \boxed{0.0121 \text{ A} = 12.1 \text{ mA}}$

Direction (Lenz's law):

B is increasing into page
Induced current opposes this change
Induced B must point out of page
By right-hand rule: current flows counterclockwise (viewed from front)

(c) Total charge:

$Q = \int I \, dt = \int \frac{|\mathcal{E}|}{R} dt = \frac{1}{R} \int |\mathcal{E}| \, dt$

$Q = \frac{\Delta \Phi_B}{R} = \frac{0.0121}{2.0}$

$Q = \boxed{6.03 \times 10^{-3} \text{ C} = 6.03 \text{ mC}}$

2Problem 2medium

❓ Question:

💡 Show Solution

Given:

r = 0.08 m
R = 2.0 Ω
B₀ = 0.2 T, B₁ = 0.8 T
Δt = 0.5 s
Loop perpendicular to field

(a) Induced EMF:

Area: $A = \pi r^2 = \pi(0.08)^2 = 0.0201$ m²

Change in flux: $\Delta \Phi_B = A \Delta B = A(B_1 - B_0)$ $\Delta \Phi_B = (0.0201)(0.8 - 0.2) = 0.0121 \text{ Wb}$

Faraday's law: $|\mathcal{E}| = \left|\frac{\Delta \Phi_B}{\Delta t}\right| = \frac{0.0121}{0.5}$

$|\mathcal{E}| = \boxed{0.0241 \text{ V} = 24.1 \text{ mV}}$

(b) Induced current:

$I = \frac{|\mathcal{E}|}{R} = \frac{0.0241}{2.0}$

$I = \boxed{0.0121 \text{ A} = 12.1 \text{ mA}}$

Direction (Lenz's law):

B is increasing into page
Induced current opposes this change
Induced B must point out of page
By right-hand rule: current flows counterclockwise (viewed from front)

(c) Total charge:

$Q = \int I \, dt = \int \frac{|\mathcal{E}|}{R} dt = \frac{1}{R} \int |\mathcal{E}| \, dt$

$Q = \frac{\Delta \Phi_B}{R} = \frac{0.0121}{2.0}$

$Q = \boxed{6.03 \times 10^{-3} \text{ C} = 6.03 \text{ mC}}$

3Problem 3hard

❓ Question:

A rectangular loop (0.5 m × 0.3 m) moves at constant velocity v = 2.0 m/s into a region where B = 0.6 T (perpendicular to loop). The loop has resistance R = 0.8 Ω. At the instant when the loop is partially in the field (0.2 m inside): Find (a) the induced EMF, (b) the induced current and power dissipated, and (c) the magnetic force on the loop.

💡 Show Solution

Given:

Dimensions: 0.5 m (width) × 0.3 m (height)
v = 2.0 m/s
B = 0.6 T
R = 0.8 Ω
x = 0.2 m inside field

(a) Induced EMF:

As loop enters field, flux through it increases: $\Phi_B = B \cdot (x \times \text{height}) = B \cdot x \cdot h$

Rate of change: $\frac{d\Phi_B}{dt} = Bh\frac{dx}{dt} = Bhv$

Motional EMF: $|\mathcal{E}| = Bhv = (0.6)(0.3)(2.0)$

$|\mathcal{E}| = \boxed{0.36 \text{ V}}$

Alternatively: $\mathcal{E} = Blv$ where l = height

(b) Current and power:

$I = \frac{|\mathcal{E}|}{R} = \frac{0.36}{0.8}$

$I = \boxed{0.45 \text{ A}}$

Direction: By Lenz's law, current opposes flux increase

Induced current creates field out of page (if B into page)
Current flows counterclockwise in loop

Power dissipated: $P = I^2 R = (0.45)^2(0.8) = \boxed{0.162 \text{ W}}$

Or: $P = \mathcal{E}I = (0.36)(0.45) = 0.162$ W ✓

(c) Magnetic force:

Force on current-carrying edge in field: $F = BIl = (0.6)(0.45)(0.3)$

$F = \boxed{0.081 \text{ N}}$

Direction: By Lenz's law, force opposes motion

Force points opposite to velocity (to the left)
This is the force you must overcome to maintain constant v

Check energy: $Fv = (0.081)(2.0) = 0.162$ W = P ✓

4Problem 4hard

❓ Question:

💡 Show Solution

Given:

Dimensions: 0.5 m (width) × 0.3 m (height)
v = 2.0 m/s
B = 0.6 T
R = 0.8 Ω
x = 0.2 m inside field

(a) Induced EMF:

As loop enters field, flux through it increases: $\Phi_B = B \cdot (x \times \text{height}) = B \cdot x \cdot h$

Rate of change: $\frac{d\Phi_B}{dt} = Bh\frac{dx}{dt} = Bhv$

Motional EMF: $|\mathcal{E}| = Bhv = (0.6)(0.3)(2.0)$

$|\mathcal{E}| = \boxed{0.36 \text{ V}}$

Alternatively: $\mathcal{E} = Blv$ where l = height

(b) Current and power:

$I = \frac{|\mathcal{E}|}{R} = \frac{0.36}{0.8}$

$I = \boxed{0.45 \text{ A}}$

Direction: By Lenz's law, current opposes flux increase

Induced current creates field out of page (if B into page)
Current flows counterclockwise in loop

Power dissipated: $P = I^2 R = (0.45)^2(0.8) = \boxed{0.162 \text{ W}}$

Or: $P = \mathcal{E}I = (0.36)(0.45) = 0.162$ W ✓

(c) Magnetic force:

Force on current-carrying edge in field: $F = BIl = (0.6)(0.45)(0.3)$

$F = \boxed{0.081 \text{ N}}$

Direction: By Lenz's law, force opposes motion

Force points opposite to velocity (to the left)
This is the force you must overcome to maintain constant v

Check energy: $Fv = (0.081)(2.0) = 0.162$ W = P ✓

5Problem 5hard

❓ Question:

A long solenoid (n = 2000 turns/m, radius R = 0.04 m) has current I(t) = I₀sin(ωt) where I₀ = 5.0 A and ω = 100π rad/s. A single circular loop of radius r = 0.06 m and resistance R = 0.5 Ω is placed around the solenoid. Find: (a) the magnetic flux through the loop as a function of time, (b) the maximum induced EMF, and (c) the maximum power dissipated.

💡 Show Solution

Given:

Solenoid: n = 2000 turns/m, R_s = 0.04 m
I(t) = I₀sin(ωt), I₀ = 5.0 A, ω = 100π rad/s
Loop: r = 0.06 m, R = 0.5 Ω
μ₀ = 4π × 10⁻⁷ T·m/A

(a) Flux through loop:

B inside solenoid: $B(t) = \mu_0 n I(t) = \mu_0 n I_0 \sin(\omega t)$

Since r > R_s, only the solenoid cross-section contributes: $\Phi_B(t) = B(t) \cdot \pi R_s^2 = \mu_0 n I_0 \pi R_s^2 \sin(\omega t)$

$\Phi_B(t) = (4\pi \times 10^{-7})(2000)(5.0)(\pi)(0.04)^2 \sin(100\pi t)$

$\Phi_B(t) = (5.03 \times 10^{-5}) \sin(100\pi t) \text{ Wb}$

$\boxed{\Phi_B(t) = 5.03 \times 10^{-5} \sin(100\pi t) \text{ Wb}}$

(b) Maximum induced EMF:

$\mathcal{E}(t) = -\frac{d\Phi_B}{dt} = -\Phi_0 \omega \cos(\omega t)$

where $\Phi_0 = 5.03 \times 10^{-5}$ Wb

$|\mathcal{E}_{max}| = \Phi_0 \omega = (5.03 \times 10^{-5})(100\pi)$

$|\mathcal{E}_{max}| = \boxed{0.0158 \text{ V} = 15.8 \text{ mV}}$

(c) Maximum power:

Maximum current: $I_{max} = \frac{|\mathcal{E}_{max}|}{R} = \frac{0.0158}{0.5} = 0.0316 \text{ A}$

$P_{max} = I_{max}^2 R = (0.0316)^2(0.5)$

$P_{max} = \boxed{5.0 \times 10^{-4} \text{ W} = 0.50 \text{ mW}}$

Or: $P_{max} = \frac{\mathcal{E}_{max}^2}{R} = \frac{(0.0158)^2}{0.5} = 5.0 \times 10^{-4}$ W ✓

6Problem 6hard

❓ Question:

💡 Show Solution

Given:

Solenoid: n = 2000 turns/m, R_s = 0.04 m
I(t) = I₀sin(ωt), I₀ = 5.0 A, ω = 100π rad/s
Loop: r = 0.06 m, R = 0.5 Ω
μ₀ = 4π × 10⁻⁷ T·m/A

(a) Flux through loop:

B inside solenoid: $B(t) = \mu_0 n I(t) = \mu_0 n I_0 \sin(\omega t)$

Since r > R_s, only the solenoid cross-section contributes: $\Phi_B(t) = B(t) \cdot \pi R_s^2 = \mu_0 n I_0 \pi R_s^2 \sin(\omega t)$

$\Phi_B(t) = (4\pi \times 10^{-7})(2000)(5.0)(\pi)(0.04)^2 \sin(100\pi t)$

$\Phi_B(t) = (5.03 \times 10^{-5}) \sin(100\pi t) \text{ Wb}$

$\boxed{\Phi_B(t) = 5.03 \times 10^{-5} \sin(100\pi t) \text{ Wb}}$

(b) Maximum induced EMF:

$\mathcal{E}(t) = -\frac{d\Phi_B}{dt} = -\Phi_0 \omega \cos(\omega t)$

where $\Phi_0 = 5.03 \times 10^{-5}$ Wb

$|\mathcal{E}_{max}| = \Phi_0 \omega = (5.03 \times 10^{-5})(100\pi)$

$|\mathcal{E}_{max}| = \boxed{0.0158 \text{ V} = 15.8 \text{ mV}}$

(c) Maximum power:

Maximum current: $I_{max} = \frac{|\mathcal{E}_{max}|}{R} = \frac{0.0158}{0.5} = 0.0316 \text{ A}$

$P_{max} = I_{max}^2 R = (0.0316)^2(0.5)$

$P_{max} = \boxed{5.0 \times 10^{-4} \text{ W} = 0.50 \text{ mW}}$

Or: $P_{max} = \frac{\mathcal{E}_{max}^2}{R} = \frac{(0.0158)^2}{0.5} = 5.0 \times 10^{-4}$ W ✓

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