Question

A 10-cm-diameter parallel-plate capacitor has a 1.0 mm spacing. the electric field between the plates is increasing at the rate 1.0 * 106 v/m s. what is the magnetic field strength (a) on the axis, (b) 3.0 cm from the axis, and (c) 7.0 cm from the axis?

Answer 1

(a) 0

The magnetic field strength insidea a parallel-plate capacitor with changing electric field can be found by applying Ampere's law:

$(2\pi r) B = \mu_0 I_D$ (1)

where

$(2\pi r)$ is the circumference of the circular line of radius r with axis coincident to the axis of the capacitor, used to calculate the magnetic field

B is the strength of the magnetic field

$I_D$ is the displacement current enclosed by the area of the circular line mentioned above, and it is equal to

$I_D = \epsilon_0 \frac{d\Phi_E}{dt} = \epsilon_0 (\pi r^2) \frac{dE}{dt}$ (2)

where

$\frac{d\Phi_E}{dt}$ is the rate of change of electric flux through the area enclosed by the line

$\frac{dE}{dt}=1.0\cdot 10^6 V/m$ is the rate of change of the electric field

Rewriting eq.(1), we find

$B = \frac{\mu_0 \epsilon_0 r}{2}\frac{dE}{dt}$

which is valid for r < R (where R=5.0 cm is the radius of the plates of the capacitor).

In this part of the problem,

r = 0

since we are on the axis; so substituting r=0 inside the formula above, we find

B(0) = 0

(b) $1.67\cdot 10^{-13}T$

In this part, we have

r = 3.0 cm = 0.03 m

The formula used in part (a) is still valid since r<R, so we can directly use it to find the magnitude of the magnetic field:

$B = \frac{\mu_0 \epsilon_0 r}{2}\frac{dE}{dt}=\frac{(4\pi\cdot 10^{-7}H/m)(8.85\cdot 10^{-12}F/m)(0.03 m)}{2}(1.0\cdot 10^6 V/m)=1.67\cdot 10^{-13}T$

(c) $1.98\cdot 10^{-13} T$

In this part, we have

r = 7.0 cm = 0.07 m

so here

r > R

therefore we need to substitute $(\pi r^2)$ with $(\pi R^2)$ in eq. (2), since the area through which the flux is calculated is only $(\pi R^2)$ (there is no electric field outside the area of the capacitor). So we find

$I_D = \epsilon_0 (\pi R^2) \frac{dE}{dt}$

and therefore

$B = \frac{\mu_0 \epsilon_0 R^2}{2r}\frac{dE}{dt}=\frac{(4\pi\cdot 10^{-7}H/m)(8.85\cdot 10^{-12}F/m)(0.05 m)^2}{2(0.07 m)}(1.0\cdot 10^6 V/m)=1.98\cdot 10^{-13} T$