To solve this problem, we need to calculate the magnetic field within the capacitor at a specific radial distance due to the displacement current as the capacitor charges. Let's break down the steps:

Given dataTo solve this problem, we need to calculate the magnetic field within the capacitor at a specific radial distance due to the displacement current as the capacitor charges. Let's break down the steps:

Given

• Battery voltage $V = 5.80 \, \text{V}$
• Resistor $R = 10.0 \, \text{M}\Omega = 10.0 \times 10^6 \, \Omega$
• Capacitor separation $d = 3.50 \, \text{mm} = 3.50 \times 10^{-3} \, \text{m}$
• Radius of capacitor plates $a = 4.00 \, \text{cm} = 0.04 \, \text{m}$
• Time $t = 170 \, \mu\text{s} = 170 \times 10^{-6} \, \text{s}$
• Radial distance within capacitor $r = 2.20 \, \text{cm} = 0.022 \, \text{m}$

Approach:

1. Calculate the capacitance of the parallel-plate capacitor: $C = \frac{\epsilon_0 \cdot A}{d}$ where $A = \pi a^2$ is the area of the plates and $\epsilon_0 = 8.854 \times 10^{-12} \, \text{F/m}$.

2. Find the current in the RC circuit, which decreases over time as the capacitor charges. The current at any time $t$ is given by: $I(t) = \frac{V}{R} \cdot e^{-\frac{t}{RC}}$

3. Calculate the displacement current $I_d$, which equals the current $I(t)$ because the electric field changes as the capacitor charges.

4. Determine the magnetic field at the distance $r$ using Ampère's law for displacement current: $B \cdot 2 \pi r = \mu_0 I_d \frac{r^2}{a^2}$ Solving for $B$: $B = \frac{\mu_0 I_d r}{2 \pi a^2}$ where $\mu_0 = 4 \pi \times 10^{-7} \, \text{T}\cdot\text{m/A}$.

Step-by-Step Solution:

Step 1: Calculate Capacitance

$C = \frac{\epsilon_0 \cdot \pi a^2}{d} = \frac{(8.854 \times 10^{-12}) \cdot \pi \cdot (0.04)^2}{3.5 \times 10^{-3}}$

Step 2: Calculate Initial Current

$I_0 = \frac{V}{R} = \frac{5.80}{10.0 \times 10^6}$

Step 3: Calculate Current at $t = 170 \, \mu\text{s}$

$I(t) = I_0 \cdot e^{-\frac{t}{RC}}$

Step 4: Calculate Magnetic Field

Substitute $I_d = I(t)$ into the equation for $B$: $B = \frac{\mu_0 I_d r}{2 \pi a^2}$

I’ll now compute each step for the final result.The magnitude of the magnetic field within the capacitor at a radial distance of 2.20 cm is approximately $4.19 \times 10^{-13} \, \text{T}$ (teslas).

Would you like further details on any part of this solution, or do you have additional questions?

Here are some related questions you might find interesting:

1. How does the displacement current relate to the real current in this RC circuit?
2. What would happen to the magnetic field if the radial distance $r$ was closer to or further from the center?
3. How does the magnetic field change over time as the capacitor charges?
4. What is the physical significance of the displacement current in a capacitor?
5. How would changing the resistor's value affect the charging time and magnetic field?

Tip: In circuits with capacitors, the displacement current can be as significant as the real current when analyzing the magnetic effects within the capacitor plates.