The image presents a series of physics problems related to circuits involving capacitors, inductors, and resistors (RL and LC circuits). We'll solve part V regarding the solenoid: a. Calculate the inductance of the solenoid. b. Calculate the resistance of the wire. c. Calculate the time constant of the RL circuit. d. Calculate the magnetic field inside the solenoid. e. Calculate the magnetic energy stored in the solenoid. Given: solenoid length $l = 62.8 \text{ cm} = 0.628 \text{ m}$, diameter $d = 8 \text{ cm} = 0.08 \text{ m}$, wire diameter $d_w = 1.256 \text{ mm} = 1.256 \times 10^{-3} \text{ m}$, voltage $V = 3.2 \text{ V}$, permeability of free space $\mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot \text{m/A}$, resistivity $\rho = 1.6 \mu\Omega\cdot \text{cm} = 1.6 \times 10^{-8} \Omega\cdot \text{m}$.

1. Problem Description

The image presents a series of physics problems related to circuits involving capacitors, inductors, and resistors (RL and LC circuits). We'll solve part V regarding the solenoid:

a. Calculate the inductance of the solenoid.

b. Calculate the resistance of the wire.

c. Calculate the time constant of the RL circuit.

d. Calculate the magnetic field inside the solenoid.

e. Calculate the magnetic energy stored in the solenoid.

Given: solenoid length $l = 62.8 \text{ cm} = 0.628 \text{ m}$, diameter $d = 8 \text{ cm} = 0.08 \text{ m}$, wire diameter $d_w = 1.256 \text{ mm} = 1.256 \times 10^{-3} \text{ m}$, voltage $V = 3.2 \text{ V}$, permeability of free space $\mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot \text{m/A}$, resistivity $\rho = 1.6 \mu\Omega\cdot \text{cm} = 1.6 \times 10^{-8} \Omega\cdot \text{m}$.

2. Solution Steps

a. Calculate the inductance of the solenoid.

First, find the number of turns $N$. Since the wires are wound tightly next to each other, the number of turns is the length of the solenoid divided by the diameter of the wire:

$N = \frac{l}{d_w} = \frac{0.628}{1.256 \times 10^{-3}} = 500$.

The inductance $L$ of a solenoid is given by:

$L = \frac{\mu_0 N^2 A}{l}$, where $A = \pi r^2$ is the cross-sectional area and $r = d/2$ is the radius.

$r = \frac{0.08}{2} = 0.04 \text{ m}$

$A = \pi (0.04)^2 = \pi (0.0016) \approx 0.0050265 \text{ m}^2$

$L = \frac{(4\pi \times 10^{-7}) (500)^2 (0.0050265)}{0.628} = \frac{4\pi \times 10^{-7} \times 250000 \times 0.0050265}{0.628} = \frac{0.001579}{0.628} \approx 0.002514 \text{ H} = 2.514 \text{ mH}$

b. Calculate the resistance of the wire.

First, find the length of the wire $l_w$ used to wind the solenoid:

$l_w = N \times 2\pi r = 500 \times 2\pi (0.04) = 500 \times 0.2513 = 125.66 \text{ m}$

The resistance $R$ of the wire is given by:

$R = \frac{\rho l_w}{A_w}$, where $A_w$ is the cross-sectional area of the wire.

$A_w = \pi (d_w/2)^2 = \pi (1.256 \times 10^{-3}/2)^2 = \pi (0.628 \times 10^{-3})^2 = \pi (3.94384 \times 10^{-7}) \approx 1.2397 \times 10^{-6} \text{ m}^2$

$R = \frac{(1.6 \times 10^{-8})(125.66)}{1.2397 \times 10^{-6}} = \frac{2.01056 \times 10^{-6}}{1.2397 \times 10^{-6}} \approx 1.6218 \Omega$

c. Calculate the time constant of the RL circuit.

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

$\tau = \frac{L}{R} = \frac{0.002514}{1.6218} \approx 0.00155 \text{ s} = 1.55 \text{ ms}$

d. Calculate the magnetic field inside the solenoid.

The magnetic field $B$ inside a solenoid is given by:

$B = \mu_0 \frac{N}{l} I$

First, find the current $I$ using Ohm's law: $V = IR$, so $I = \frac{V}{R} = \frac{3.2}{1.6218} \approx 1.973 \text{ A}$

$B = (4\pi \times 10^{-7}) \frac{500}{0.628} (1.973) = (4\pi \times 10^{-7}) (796.178) (1.973) \approx 1.968 \times 10^{-3} \text{ T} = 1.968 \text{ mT}$

e. Calculate the magnetic energy stored in the solenoid.

The magnetic energy $U$ stored in an inductor is given by:

$U = \frac{1}{2} LI^2 = \frac{1}{2} (0.002514) (1.973)^2 = \frac{1}{2} (0.002514) (3.892729) \approx 0.00489 \text{ J} = 4.89 \text{ mJ}$

3. Final Answer

a. Inductance: $L = 2.514 \text{ mH}$

b. Resistance: $R = 1.6218 \Omega$

c. Time constant: $\tau = 1.55 \text{ ms}$

d. Magnetic field: $B = 1.968 \text{ mT}$

e. Magnetic energy: $U = 4.89 \text{ mJ}$