A parallel plate capacitor with circular plates of radius 10.0 cm are separated by a distance of 5.0 mm.The potential across the capacitor is increased at a constant rate of 120 V/s.Determine the magnitude of the magnetic field between the plates at a distance r = 1.0 cm from the center.

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Quizplus · Accepted Answer

The magnetic field between the plates of a parallel plate capacitor, when the potential across the capacitor is increasing, can be determined using the concept of displacement current. The displacement current $I_d$ can be found using the equation $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$, where $\epsilon_0$ is the permittivity of free space, and $\frac{d\Phi_E}{dt}$ is the rate of change of electric flux. Since the electric flux $\Phi_E = EA = \frac{V}{d} \cdot \pi r^2$ for a parallel plate capacitor, where $V$ is the potential difference, $d$ is the separation between the plates, and $r$ is the radius at which we're calculating the magnetic field, the rate of change of $V$ with time ($\frac{dV}{dt}$) is given as 120 V/s. The magnetic field at a distance $r$ from the center due to the displacement current can be found using Ampère's law, which in this context is $B \cdot 2\pi r = \mu_0 I_d$, where $B$ is the magnetic field and $\mu_0$ is the permeability of free space.Given:- $r = 1.0 \, \text{cm} = 0.01 \, \text{m}$- $\frac{dV}{dt} = 120 \, \text{V/s}$- $d = 5.0 \, \text{mm} = 0.005 \, \text{m}$- $\epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m}$- $\mu_0 = 4\pi \times 10^{-7} \, \text{Tm/A}$The calculation involves determining the displacement current and then using it to find the magnetic field. The correct answer involves applying these principles correctly to find the magnitude of the magnetic field, which is calculated to be $1.3 \times 10^{-15} \, \text{T}$, indicating a very small magnetic field strength at the specified location between the plates.

A Parallel Plate Capacitor with Circular Plates of Radius 10