The image contains several physics problems. I will focus on problem IV. A long solenoid has a length of $8.0 \, \text{cm}$ and $N = 500$ turns. The current through the solenoid changes from $0 \, \text{A}$ to $2.5 \, \text{A}$ in $0.35 \, \text{s}$. An induced emf of $0.012 \, \text{V}$ is generated. We need to calculate: a. The inductance of the solenoid. b. The cross-sectional area of the solenoid.

1. Problem Description

The image contains several physics problems. I will focus on problem IV.

A long solenoid has a length of $8.0 \, \text{cm}$ and $N = 500$ turns. The current through the solenoid changes from $0 \, \text{A}$ to $2.5 \, \text{A}$ in $0.35 \, \text{s}$. An induced emf of $0.012 \, \text{V}$ is generated. We need to calculate:

a. The inductance of the solenoid.

b. The cross-sectional area of the solenoid.

2. Solution Steps

a. Calculating the inductance of the solenoid:

The induced emf $\epsilon$ is related to the inductance $L$ and the rate of change of current $\frac{dI}{dt}$ by the formula:

$$\epsilon = -L \frac{dI}{dt}$$

We are given that $\epsilon = 0.012 \, \text{V}$, $\frac{dI}{dt} = \frac{2.5 \, \text{A} - 0 \, \text{A}}{0.35 \, \text{s}} = \frac{2.5}{0.35} \, \text{A/s}$.

So, we have:

$$0.012 = L \times \frac{2.5}{0.35}$$

$$L = \frac{0.012 \times 0.35}{2.5} = \frac{0.0042}{2.5} = 0.00168 \, \text{H}$$

$$L = 1.68 \, \text{mH}$$

b. Calculating the cross-sectional area of the solenoid:

The inductance of a solenoid is given by:

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

where $L$ is the inductance, $\mu_0 = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}$ is the permeability of free space, $N$ is the number of turns, $A$ is the cross-sectional area, and $l$ is the length of the solenoid.

We have $L = 0.00168 \, \text{H}$, $N = 500$, $l = 8.0 \, \text{cm} = 0.08 \, \text{m}$.

So,

$$0.00168 = \frac{4\pi \times 10^{-7} \times 500^2 \times A}{0.08}$$

$$A = \frac{0.00168 \times 0.08}{4\pi \times 10^{-7} \times 250000} = \frac{0.0001344}{0.314159} = 4.278 \times 10^{-7} \, \text{m}^2$$

$$A \approx 4.28 \times 10^{-7} \, \text{m}^2$$

3. Final Answer

a. The inductance of the solenoid is $1.68 \, \text{mH}$.

b. The cross-sectional area of the solenoid is $4.28 \times 10^{-7} \, \text{m}^2$.