Question

A parallel plate capacitor has circular plates of radius \(10.0 \mathrm{~cm}\) that are separated by a distance of \(5.00 \mathrm{~mm}\). The potential across the capacitor is increased at a constant rate of \(1.20 \mathrm{kV} / \mathrm{s}\). Determine the magnitude of the magnetic field between the plates at a distance \(r=4.00 \mathrm{~cm}\) from the center.

Short answer

Answer: The magnitude of the magnetic field between the plates at a distance r = 4.00 cm from the center is approximately 1.34 x 10^{-9} T.

Step-by-step solution

1. Write down the Ampere-Maxwell Law

The Ampere-Maxwell Law is given by the equation: $$ \oint \vec{B} \cdot \mathrm{d}\vec{l} = \mu_0 (I + \epsilon_0 \frac{\mathrm{d}\Phi_E}{\mathrm{d}t})$$ In this case, we don't have any current flowing through the wire, so \(I = 0\). Therefore, the equation becomes: $$ \oint \vec{B} \cdot \mathrm{d}\vec{l} = \epsilon_0 \mu_0 \frac{\mathrm{d}\Phi_E}{\mathrm{d}t} $$

2. Get an expression for the magnetic field B and displacement current density J

Since we are interested in the magnetic field at a distance r from the center of the plates, we can use symmetry and the fact that the magnetic field will be tangential to the circular path of radius r. Therefore, we can write: $$ B \cdot 2\pi r = \epsilon_0 \mu_0 \frac{\mathrm{d}\Phi_E}{\mathrm{d}t} $$ The displacement current density J is given by the rate of change of the electric field, so we need to express the electric field in terms of the change in potential, and the change in potential is given in the problem as 1.20 kV/s. We can write the electric field relation as: $$ E = \frac{V}{d} $$ Where d is the distance between the plates (5.00 mm). The rate of change of the electric field is: $$ \frac{\mathrm{d}E}{\mathrm{d}t} = \frac{1}{d} \frac{\mathrm{d}V}{\mathrm{d}t} $$

3. Relate the magnetic field B to the displacement current density J

Now we need to relate the rate of change of the electric field to the displacement current density using the formula: $$ J = \epsilon_0 \frac{\mathrm{d}E}{\mathrm{d}t} $$ Substitute the expression for the rate of change of the electric field: $$ J = \epsilon_0 \frac{1}{d} \frac{\mathrm{d}V}{\mathrm{d}t} $$ We can now relate the displacement current density J to the magnetic field B using the formula: $$ B \cdot 2\pi r = \epsilon_0 \mu_0 J \cdot Area $$ Where Area is the area of the circular path of radius r: $$ Area = \pi r^2 $$

4. Calculate the magnetic field B

From the previous step, we can isolate the magnetic field B and substitute for J: $$ B = \frac{\epsilon_0 \mu_0 \cdot (\epsilon_0 \frac{1}{d} \frac{\mathrm{d}V}{\mathrm{d}t}) \cdot \pi r^2}{2\pi r} $$ We can now plug in the given values: $$ \epsilon_0 = 8.85 \times 10^{-12} \, \mathrm{F/m} $$ $$ \mu_0 = 4\pi \times 10^{-7} \, \mathrm{T \cdot m/A} $$ $$ r = 4 \times 10^{-2} \, \mathrm{m} $$ $$ d = 5 \times 10^{-3} \, \mathrm{m} $$ $$ \frac{\mathrm{d}V}{\mathrm{d}t} = 1.20 \times 10^3 \, \mathrm{V/s} $$ Plug in these values into the equation for B and get the magnitude of the magnetic field: $$ B = \frac{(8.85 \times 10^{-12} \mathrm{F/m}) \cdot (4\pi \times 10^{-7} \mathrm{T \cdot m/A}) \cdot (\frac{1}{5 \times 10^{-3} \mathrm{m}}) (1.20 \times 10^3 \mathrm{V/s}) \cdot \pi (4 \times 10^{-2} \mathrm{m})^2}{2\pi (4 \times 10^{-2} \mathrm{m})} $$ Calculate B: $$ B \approx 1.34 \times 10^{-9} \, \mathrm{T} $$ The magnitude of the magnetic field between the plates at a distance r = 4.00 cm from the center is approximately 1.34 x 10^{-9} T.