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: To determine the magnitude of the magnetic field at a distance of 4.0 cm from the center, we can follow these steps: (1) calculate the capacitance of the capacitor, (2) calculate the displacement current, (3) use Ampere's law with Maxwell's addition, and (4) determine the magnetic field. By doing the calculations, we can find the magnitude of the magnetic field between the plates at the given distance.

Step-by-step solution

Calculate the capacitance of the capacitor

To calculate the capacitance of the parallel plate capacitor, use the following formula: \(C = \frac{\epsilon_0 A}{d}\) where \(C\) is the capacitance, \(\epsilon_0\) is the vacuum permittivity (\(8.85\times 10^{-12} \mathrm{F/m}\)), \(A\) is the area of the plates, and \(d\) is the distance between the plates. Here, \(A = \pi r^2 = \pi (0.1)^2\) and \(d = 5.0 \times 10^{-3} \mathrm{m}\).

Calculate the displacement current

Since the potential across the capacitor is increasing at a constant rate of \(1200 \mathrm{V/s}\), we can calculate the rate of change of charge on the capacitor using the following equation: \(\frac{dQ}{dt} = C\frac{dV}{dt}\) where \(\frac{dQ}{dt}\) is the rate of change of the charge and \(\frac{dV}{dt}\) is the rate of change of the voltage. Using the capacitance (\(C\)) calculated in Step 1, we can find the rate of change of charge on the capacitor.

Use Ampere's law with Maxwell's addition

To find the magnetic field at a distance \(r = 4.0 \mathrm{cm}\) from the center, we can use Ampere's law with Maxwell's addition: \(\oint Bdl = \mu_0\big(I_c + \epsilon_0\frac{d\Phi_E}{dt}\big)\) where \(B\) is the magnetic field, \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7} \mathrm{Tm/A}\)), \(I_c\) is the conduction current (which is zero in this case), and \(\epsilon_0\frac{d\Phi_E}{dt}\) is the displacement current. In this case, the electric flux \(\Phi_E = E \cdot A\) and the displacement current can be written as: \(\epsilon_0\frac{d\Phi_E}{dt} = \epsilon_0\frac{d(EA)}{dt} = \epsilon_0 A\frac{dE}{dt}\) Since we have a parallel plate capacitor, we can say \(E = \frac{V}{d}\), so the displacement current becomes: \(\epsilon_0 A\frac{dE}{dt} = \epsilon_0 A\frac{d(\frac{V}{d})}{dt} = \epsilon_0 \frac{A}{d}\frac{dV}{dt} = C\frac{dV}{dt}\) Now, we can use the integral form of Ampere's law with Maxwell's addition to find the magnetic field.

Determine the magnetic field

As the magnetic field will be in the form of circular loops, we can use the integral form of Ampere's law with Maxwell's addition as: \(B\oint dl = B(2\pi r) = \mu_0C\frac{dV}{dt}\) Then, to find the magnitude of the magnetic field between the plates at a distance \(r=4.0 \mathrm{cm}\) from the center, we can rearrange the equation: \(B = \frac{\mu_0C\frac{dV}{dt}}{2\pi r}\) Using the values calculated in the previous steps, we can find the magnitude of the magnetic field at the given distance.