The problem describes a solenoid with the following parameters: Length $l = 62.8 cm = 0.628 m$ Diameter $d = 8 cm = 0.08 m$, so the radius is $r = 0.04 m$ Wire diameter $d_{wire} = 1.256 mm = 1.256 \times 10^{-3} m$ Voltage across the solenoid $V = 3.2 V$ Permeability of free space $\mu_0 = 4\pi \times 10^{-7} T \cdot m/A$ Resistivity of the wire $\rho = 1.6 \mu \Omega \cdot cm = 1.6 \times 10^{-8} \Omega \cdot m$ The problem asks to calculate the following: a. Inductance of the solenoid ($L$) b. Resistance of the wire ($R$) c. Time constant of the RL circuit ($\tau$) d. Magnetic flux density inside the solenoid ($B$) e. Magnetic energy stored in the solenoid ($U$)

1. Problem Description

The problem describes a solenoid with the following parameters:

Length $l = 62.8 cm = 0.628 m$

Diameter $d = 8 cm = 0.08 m$, so the radius is $r = 0.04 m$

Wire diameter $d_{wire} = 1.256 mm = 1.256 \times 10^{-3} m$

Voltage across the solenoid $V = 3.2 V$

Permeability of free space $\mu_0 = 4\pi \times 10^{-7} T \cdot m/A$

Resistivity of the wire $\rho = 1.6 \mu \Omega \cdot cm = 1.6 \times 10^{-8} \Omega \cdot m$

The problem asks to calculate the following:

a. Inductance of the solenoid ($L$)

b. Resistance of the wire ($R$)

c. Time constant of the RL circuit ($\tau$)

d. Magnetic flux density inside the solenoid ($B$)

e. Magnetic energy stored in the solenoid ($U$)

2. Solution Steps

a. Inductance of the solenoid ($L$)

The inductance of a solenoid is given by:

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

where $N$ is the number of turns, $A$ is the cross-sectional area, and $l$ is the length.

The area is $A = \pi r^2 = \pi (0.04)^2 = 0.0016\pi m^2$.

The number of turns can be estimated from the length of the solenoid and the diameter of the wire. Since the turns are close together, the number of turns $N$ is roughly $l/d_{wire}$.

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

Then,

$L = \frac{(4\pi \times 10^{-7}) (500)^2 (0.0016\pi)}{0.628} = \frac{(4\pi \times 10^{-7}) (250000) (0.0016\pi)}{0.628} \approx 7.9 \times 10^{-4} H = 0.79 mH$

b. Resistance of the wire ($R$)

We know that the voltage across the solenoid is $3.2 V$. We need to find the current. We'll calculate the total length of the wire.

The total length of the wire $l_{wire} = N (2 \pi r) = 500 \times 2 \pi \times 0.04 = 40\pi \approx 125.66 m$.

Resistance $R = \rho \frac{l_{wire}}{A_{wire}}$ where $A_{wire} = \pi (d_{wire}/2)^2 = \pi (1.256 \times 10^{-3} / 2)^2 = \pi (0.628 \times 10^{-3})^2 \approx 1.239 \times 10^{-6} m^2$

$R = (1.6 \times 10^{-8}) \frac{125.66}{1.239 \times 10^{-6}} = \frac{2.01 \times 10^{-6}}{1.239 \times 10^{-6}} \approx 1.62 \Omega$

c. Time constant of the RL circuit ($\tau$)

The time constant is given by:

$\tau = \frac{L}{R} = \frac{7.9 \times 10^{-4}}{1.62} \approx 4.88 \times 10^{-4} s = 0.488 ms$

d. Magnetic flux density inside the solenoid ($B$)

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

We can calculate the current $I$ using Ohm's law:

$I = \frac{V}{R} = \frac{3.2}{1.62} \approx 1.98 A$

$B = (4\pi \times 10^{-7}) \frac{500}{0.628} (1.98) \approx 1.98 \times 10^{-3} T$

e. Magnetic energy stored in the solenoid ($U$)

$U = \frac{1}{2} L I^2 = \frac{1}{2} (7.9 \times 10^{-4}) (1.98)^2 \approx 1.55 \times 10^{-3} J$

3. Final Answer

a. Inductance of the solenoid: $L = 0.79 mH$

b. Resistance of the wire: $R = 1.62 \Omega$

c. Time constant of the RL circuit: $\tau = 0.488 ms$

d. Magnetic flux density inside the solenoid: $B = 1.98 \times 10^{-3} T$

e. Magnetic energy stored in the solenoid: $U = 1.55 \times 10^{-3} J$