Units: 1 henry (H) = 1 Wb/A = 1 V·s/A

Inductance of Solenoid

Long solenoid: $n$ turns/length, cross-sectional area $A$, length $l$:

$L = \mu_0 n^2 Al = \mu_0 \frac{N^2}{l}A$

Mutual Inductance

Current $I_1$ in coil 1 creates flux through coil 2:

$\mathcal{E}_2 = -M\frac{dI_1}{dt}$

Mutual inductance: $M = \frac{\Phi_{21}}{I_1} = \frac{\Phi_{12}}{I_2}$

(Same value both ways: $M_{12} = M_{21}$)

Energy Stored in Inductor

$U_L = \frac{1}{2}LI^2$

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

For solenoid: $U = u_B \cdot \text{volume} = \frac{B^2}{2\mu_0}Al = \frac{1}{2}LI^2$

(Using $B = \mu_0 nI$ and $L = \mu_0 n^2 Al$)

RL Circuit: Current Growth

Circuit: battery ($\mathcal{E}$), resistor ($R$), inductor ($L$) in series.

Kirchhoff's loop rule: $\mathcal{E} - IR - L\frac{dI}{dt} = 0$

Differential equation: $\frac{dI}{dt} = \frac{\mathcal{E}}{L} - \frac{R}{L}I$

Solution: $I(t) = \frac{\mathcal{E}}{R}(1 - e^{-Rt/L})$

Time constant: $\tau_L = \frac{L}{R}$

After time $\tau_L$: current reaches $(1 - 1/e) \approx 63\%$ of final value.

RL Circuit: Current Decay

Remove battery, current decays:

$L\frac{dI}{dt} + IR = 0$

Solution: $I(t) = I_0 e^{-Rt/L}$

Current decays exponentially with time constant $\tau_L = L/R$.

Energy Considerations

Current growth:

Energy from battery: $W = \int_0^\infty \mathcal{E}I \, dt = \frac{\mathcal{E}^2L}{R^2}$

Energy stored in inductor: $U_L = \frac{1}{2}L\left(\frac{\mathcal{E}}{R}\right)^2$

Energy dissipated in resistor: $U_R = \frac{1}{2}L\left(\frac{\mathcal{E}}{R}\right)^2$

(Equal amounts stored and dissipated)

LC Circuit

Inductor and capacitor (no resistance):

$L\frac{d^2Q}{dt^2} + \frac{Q}{C} = 0$

Oscillation: $Q(t) = Q_0\cos(\omega t + \phi)$

where $\omega = 1/\sqrt{LC}$ (angular frequency).

Energy oscillates between:

• Electric: $U_E = Q^2/(2C)$
• Magnetic: $U_B = LI^2/2$
• Total: $U = U_E + U_B$ = constant

LRC Circuit

With resistance, oscillations damped:

$L\frac{d^2Q}{dt^2} + R\frac{dQ}{dt} + \frac{Q}{C} = 0$

Damped oscillation (for $R < 2\sqrt{L/C}$): $Q(t) = Q_0e^{-Rt/(2L)}\cos(\omega' t)$

where $\omega' = \sqrt{\frac{1}{LC} - \frac{R^2}{4L^2}}$

Quality factor: $Q = \frac{\omega_0 L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}}$

$\tau = \boxed{0.05 \text{ s} = 50 \text{ ms}}$

(b) Current at t = 0.08 s:

For RL circuit with battery: $I(t) = I_{max}(1 - e^{-t/\tau})$

where $I_{max} = \varepsilon/R = 12/8.0 = 1.5$ A

$I(0.08) = 1.5(1 - e^{-0.08/0.05})$

$I(0.08) = 1.5(1 - e^{-1.6})$

$I(0.08) = 1.5(1 - 0.2019)$

$I(0.08) = 1.5(0.7981) = \boxed{1.20 \text{ A}}$

(c) Rate of energy storage:

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

$\frac{dU_L}{dt} = LI\frac{dI}{dt}$

From differential equation: $\varepsilon = IR + L\frac{dI}{dt}$

$\frac{dI}{dt} = \frac{\varepsilon - IR}{L}$

At t = 0.08 s: $\frac{dI}{dt} = \frac{12 - (1.20)(8.0)}{0.4} = \frac{12 - 9.6}{0.4} = 6.0 \text{ A/s}$

$\frac{dU_L}{dt} = LI\frac{dI}{dt} = (0.4)(1.20)(6.0)$

$\frac{dU_L}{dt} = \boxed{2.88 \text{ W}}$

Alternatively: $P_L = \varepsilon I - I^2 R = VI_L$ where $V_L = L\frac{dI}{dt}$