https://file.aiquickdraw.com/mathbot/file/1730774247686wp3r27u0.jpg

To solve this problem, let's go through it step-by-step.

### Given Data:
1. **Voltage (\( V \))**: 4.70 V
2. **Resistance (\( R \))**: 50.0 MΩ (Megaohms) = \( 50.0 \times 10^6 \) Ω
3. **Capacitor's Plate Radius (\( r \))**: 4.90 cm = 0.049 m
4. **Plate Separation Distance (\( d \))**: 3.00 mm = 0.003 m
5. **Time (\( t \))**: 280 µs = \( 280 \times 10^{-6} \) s
6. **Radial Distance for Magnetic Field (\( r \))**: 3.00 cm = 0.03 m

We are asked to find the **magnitude of the magnetic field within the capacitor** at this radial distance of 3.00 cm.

### Steps to Solve:

1. **Calculate the Capacitance of the Capacitor**:
   \[
   C = \frac{\varepsilon_0 \cdot A}{d}
   \]
   where:
   - \( \varepsilon_0 \) is the permittivity of free space, \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \),
   - \( A \) is the area of one of the plates, \( A = \pi r^2 \).

   Substituting values:
   \[
   A = \pi (0.049)^2 \approx 7.54 \times 10^{-3} \, \text{m}^2
   \]
   \[
   C = \frac{(8.85 \times 10^{-12}) \cdot (7.54 \times 10^{-3})}{0.003} \approx 2.226 \times 10^{-11} \, \text{F}
   \]

2. **Calculate the Charging Current, \( I(t) \)**:
   For an RC charging circuit, the current \( I(t) \) is given by:
   \[
   I(t) = \frac{V}{R} e^{-\frac{t}{RC}}
   \]

   Calculate the time constant \( \tau = RC \):
   \[
   \tau = (50 \times 10^6) \cdot (2.226 \times 10^{-11}) \approx 1.113 \, \text{s}
   \]

   Now, calculate \( I(t) \) at \( t = 280 \, \mu\text{s} = 280 \times 10^{-6} \, \text{s} \):
   \[
   I(280 \, \mu\text{s}) = \frac{4.7}{50 \times 10^6} e^{-\frac{280 \times 10^{-6}}{1.113}}
   \]

   Calculate each term:
   - \( \frac{4.7}{50 \times 10^6} \approx 9.4 \times 10^{-8} \, \text{A} \),
   - \( e^{-\frac{280 \times 10^{-6}}{1.113}} \approx e^{-0.2515} \approx 0.7778 \).

   So:
   \[
   I(280 \, \mu\text{s}) \approx (9.4 \times 10^{-8}) \cdot 0.7778 \approx 7.31 \times 10^{-8} \, \text{A}
   \]

3. **Calculate the Magnetic Field \( B(r) \) Inside the Capacitor**:
   Using Ampère's law in its displacement current form:
   \[
   B(r) = \frac{\mu_0 I(t) \cdot r}{2 \pi R^2}
   \]
   where:
   - \( \mu_0 \) is the permeability of free space, \( \mu_0 = 4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A} \),
   - \( R \) is the plate radius (0.049 m).

   Substitute the values:
   \[
   B(0.03) = \frac{(4\pi \times 10^{-7}) \cdot (7.31 \times 10^{-8}) \cdot 0.03}{2 \pi (0.049)^2}
   \]

   Simplifying:
   - \( 4 \pi \times 10^{-7} \times 7.31 \times 10^{-8} \approx 9.17 \times 10^{-14} \),
   - \( 2 \pi (0.049)^2 \approx 1.51 \times 10^{-2} \).

   So:
   \[
   B(0.03) \approx \frac{9.17 \times 10^{-14} \cdot 0.03}{1.51 \times 10^{-2}} \approx 1.82 \times 10^{-13} \, \text{T}
   \]

### Final Answer:
The magnetic field at a radial distance of 3.00 cm within the capacitor is approximately:
\[
B = 1.82 \times 10^{-13} \, \text{T}
\]

Would you like further details on any step, or do you have any questions?

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Here are some related questions that might be of interest:
1. What would happen to the magnetic field if the resistance were halved?
2. How would the magnetic field change if the distance \( d \) between the capacitor plates was doubled?
3. How does the time constant \( \tau \) affect the rate of charge buildup in the capacitor?
4. What if the plate radius was reduced by half—how would that affect the field strength?
5. Why is Ampère's law relevant in the context of the displacement current within a capacitor?

**Tip:** In RC circuits, the time constant \( \tau = RC \) determines how quickly the capacitor charges or discharges.

The circuit in the figure consists of switch S, a 4.70 V ideal battery, a 50.0 MΩ resistor, and an air-filled capacitor. The capacitor has parallel circular plates of radius 4.90 cm, separated by 3.00 mm. At time t = 0, switch S is closed to begin charging the capacitor. The electric field between the plates is uniform. At t = 280 µs, what is the magnitude of the magnetic field within the capacitor, at radial distance 3.00 cm?

Explore the step-by-step solution for calculating the magnetic field inside a capacitor in an RC circuit. Given a 4.70 V battery, 50.0 MΩ resistor, and an air-filled capacitor with 4.90 cm radius plates, learn to find the magnetic field at 3.00 cm radial distance after 280 µs. This problem covers RC circuits, electric and magnetic field concepts, suitable for advanced physics students.

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