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:

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

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

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

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 ✓

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 ✓