A10-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?

(a) 0

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

(1)

where

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

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

(2)

where

is the rate of change of electric flux through the area enclosed by the line

is the rate of change of the electric field

Rewriting eq.(1), we find

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)

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:

(c)

In this part, we have

r = 7.0 cm = 0.07 m

so here

r > R

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

and therefore

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