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

1. Change B: $\Phi_B = BA\cos\theta$, vary $B$
2. Change A: Move wire to change loop area
3. 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.

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 ✓

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 ✓