$\mathcal{E} = -L\frac{dI}{dt}$

where $L$ is inductance (or self-inductance).

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

$\frac{dI}{I} = -\frac{R}{L}dt$

Integrating: $\ln I = -\frac{R}{L}t + C$

At t = 0: I = I₀, so C = ln I₀

$\ln\left(\frac{I}{I_0}\right) = -\frac{R}{L}t$

$\boxed{I(t) = I_0 e^{-Rt/L} = 2.0 e^{-10t/0.25} = 2.0e^{-40t} \text{ A}}$

(b) Time for I = 0.1I₀:

$0.1I_0 = I_0 e^{-Rt/L}$

$0.1 = e^{-Rt/L}$

$\ln(0.1) = -\frac{Rt}{L}$

$t = -\frac{L\ln(0.1)}{R} = -\frac{(0.25)(-2.303)}{10}$

$t = \boxed{0.0576 \text{ s} = 57.6 \text{ ms}}$

This is $2.3\tau$ where $\tau = L/R = 0.025$ s

(c) Total energy dissipated:

Initial energy stored in inductor: $U_0 = \frac{1}{2}LI_0^2 = \frac{1}{2}(0.25)(2.0)^2$

$U_0 = 0.5 \text{ J}$

As t → ∞, all energy is dissipated: $E_{dissipated} = U_0 = \boxed{0.5 \text{ J}}$

Check: $\int_0^\infty I^2 R \, dt = \int_0^\infty I_0^2 e^{-2Rt/L} R \, dt = \frac{I_0^2 R L}{2R} = \frac{1}{2}LI_0^2$ ✓

$\omega = \frac{1}{\sqrt{5.0 \times 10^{-6}}} = \frac{1}{2.24 \times 10^{-3}}$

$\omega = \boxed{447 \text{ rad/s}}$

Frequency: $f = \omega/(2\pi) = 71.2$ Hz

(b) Maximum current:

Energy conservation: $U_C = U_L$

$\frac{Q_0^2}{2C} = \frac{1}{2}LI_{max}^2$

$I_{max} = \frac{Q_0}{\sqrt{LC}} = Q_0 \omega$

$I_{max} = (1.0 \times 10^{-4})(447)$

$I_{max} = \boxed{0.0447 \text{ A} = 44.7 \text{ mA}}$

(c) Q(t) and I(t):

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

At t = 0: Q = Q₀, so φ = 0

$\boxed{Q(t) = Q_0 \cos(\omega t) = (1.0 \times 10^{-4})\cos(447t) \text{ C}}$

Current: $I(t) = -\frac{dQ}{dt} = Q_0 \omega \sin(\omega t)$

$\boxed{I(t) = I_{max} \sin(\omega t) = 0.0447\sin(447t) \text{ A}}$

Energy oscillates between capacitor and inductor:

• At $\omega t = 0, \pi, 2\pi$: all energy in C
• At $\omega t = \pi/2, 3\pi/2$: all energy in L