Question

Suppose that a circular parallel-plate capacitor has radius R0 = 3.0 cm and plate separation d = 5.0 mm. A sinusoidal potential difference V = V0 sin(2πft) is applied across the plates, where V0 = 150 V and f = 60 Hz. In the region between the plates, find the magnitude of the induced magnetic field versus R and t, where R is the radial distance from the capacitor’s central axis. Assume that B = B0(R) is the amplitude of the magnetic field.

Answer 1

Answer:

$B=\frac{\mu_o \epsilon_o R_o^2V_o}{2rd}cos(\omega t)$

Explanation:

By the information of the statement you have that the sinusoidal potential difference is given by:

$V=V_osin(\omega t)=V_osin(2\pi ft)=150sin(2\pi (60)t)$    (1)

In order to calculate the induced magnetic field in between the plates, you first take into account the following formula, which is the Ampere-Maxwell law:

$\int B\cdot ds=\mu_o \epsilon_o\frac{d\Phi_E}{dt}+\mu_oI_c$           (2)

B: induced magnetic field

μo: magnetic permeability of vacuum = 4π*10^-7 A/T

εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

Ic: conduction current

ФE: electric flux

There is no conduction current in between the plates, then Ic = 0A

Next, you calculate dФE/dt, as follow:

The electric field, and the electric flux, are:

$E=\frac{V}{d}=\frac{V_osin(\omega t)}{d}\\\\\Phi_E=EA$

d: separation between plates = 5.0mm = 5.0*10^-3 m

A: area of the circular plates = $\pi R_o^2$

Ro: radius of the circular capacitor = 3.0cm = 0.03m

Thus, dФE/dt is:

$\frac{d\Phi_E}{dt}=\frac{d(EA)}{dt}=\pi R_o^2\frac{d}{dt}[\frac{V_osin(\omega t)}{d}]\\\\\frac{d\Phi_E}{dt}=\frac{\pi \omega R_o^2 V_o}{d}cos(\omega t)$       (3)

The induced magnetic field is calculated by taking into account that the integral of the equation (2) is:

$\int B \cdot ds=B\int ds=B(2\pi r)$         (4)

Next, you replace the results of (3) and (4) into the equation (2) and you solve for B:

$B(2\pi r)=\mu_o \epsilon_o (\frac{\pi \omega R_o^2 V_o}{d}cos(\omega t))\\\\B=\frac{\mu_o \epsilon_o R_o^2V_o}{2rd}cos(\omega t)$   (5)

The last expression is de induced magnetic field in between the plates in terms of t and r

Another way of expressing the  formula (5) is as follow:

$B=B_ocos(\omega t)\\\\B_o=\frac{\mu_o \epsilon_o R_o^2V_o}{2rd}$