QuestionA 5.0 A current is charging a 0.50 -cm-diameter parallel-plate capacitor.
Part A
Part B
What is the magnetic field strength at a point 2.5 mm radially from the center of the capacitor? Express your answer with the appropriate units.

Studdy Solution
Calculate the magnetic field strength at the given point:
Substitute the values into Ampère's Law:
$$B \cdot 2\pi (2.5 \times 10^{-3} \, \text{m}) = \mu_0 \times 5.0 \, \text{A}$$
Solve for $B$:
$$B = \frac{\mu_0 \times 5.0 \, \text{A}}{2\pi \times 2.5 \times 10^{-3} \, \text{m}}$$
Using $\mu_0 = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}$:
$$B = \frac{4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A} \times 5.0 \, \text{A}}{2\pi \times 2.5 \times 10^{-3} \, \text{m}}$$
$$B = \frac{20\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}}{5\pi \times 10^{-3} \, \text{m}}$$
$$B = \frac{20 \times 10^{-7}}{5 \times 10^{-3}} \, \text{T}$$
$$B = 4 \times 10^{-4} \, \text{T}$$
The magnetic field strength at the point is:
$$\boxed{4 \times 10^{-4} \, \text{T}}$$