We are given several physics problems involving capacitors, inductors, RL circuits, and solenoids. We need to calculate various parameters like energy stored in a capacitor, current in an inductor, time constant of an RL circuit, inductance of an LC circuit, inductance of a solenoid, magnetic field inside a solenoid, and energy stored in a solenoid.

1. Problem Description

We are given several physics problems involving capacitors, inductors, RL circuits, and solenoids. We need to calculate various parameters like energy stored in a capacitor, current in an inductor, time constant of an RL circuit, inductance of an LC circuit, inductance of a solenoid, magnetic field inside a solenoid, and energy stored in a solenoid.

2. Solution Steps

II. RL Circuit Problem

a. Calculate the time constant of the RL circuit.

The time constant $\tau$ of an RL circuit is given by:

$\tau = \frac{L}{R}$

Where $L = 30.0 \, \text{mH} = 30.0 \times 10^{-3} \, \text{H}$ and $R = 6.00 \, \Omega$.

$\tau = \frac{30.0 \times 10^{-3}}{6.00} = 5.0 \times 10^{-3} \, \text{s} = 5.0 \, \text{ms}$

b. Calculate the current in the circuit at t = 2 ms.

The current in an RL circuit as a function of time is given by:

$I(t) = \frac{V}{R}(1 - e^{-\frac{t}{\tau}})$

Where $V = 12.0 \, \text{V}$, $R = 6.00 \, \Omega$, $t = 2 \, \text{ms} = 2 \times 10^{-3} \, \text{s}$, and $\tau = 5 \times 10^{-3} \, \text{s}$.

$I(2 \times 10^{-3}) = \frac{12.0}{6.00}(1 - e^{-\frac{2 \times 10^{-3}}{5 \times 10^{-3}}})$

$I(2 \times 10^{-3}) = 2(1 - e^{-0.4})$

$I(2 \times 10^{-3}) = 2(1 - 0.6703) = 2(0.3297) = 0.6594 \, \text{A}$

3. Final Answer

II. RL Circuit Problem

a. The time constant of the RL circuit is $5.0 \, \text{ms}$.

b. The current in the circuit at t = 2 ms is $0.6594 \, \text{A}$.