A circular capacitor has two metal plates of radius R=3.0 cm. The top plate has charge + Q , the bottom has charge - Q , and they are separated by a short distance. The capacitor is connected to a circuit with current I=2.5 A . Find the magnetic field strength B at a point between the plates a distance r=2.0 cm from the axis through the center of the plates.

We can find B from the generalized form of Ampere's law. Choose a circular path of radius r=2.0 cm about the center line joining the two plates, as shown in the diagram below.
Ampere's law states
oint_C B^(->)*dl^->=mu_0 (I+I_d)
Where the displacement current is I_d=epsilon_0 (d phi_e)/(dt)
From symmetry the line integral is just B multiplied by the circumference of the circle of radius r .
oint_C B^(->)*dl^(->)=B*(2pi r)
There are no charges moving through the surface S , therefore I=0 .
B*(2pi r)=mu_0I+mu_0I_d
eq. (1) :-gt B*(2pi r)=mu_0epsilon_0 (d phi_e)/(dt)
The electric flux through S equals the product of the uniform field strength E and the area A of the flat surface S bounded by the curve C , and E is equal to the surface charge sigma over epsilon_0 .
phi_e=AE=pir^2E=pir^2 sigma/epsilon_0
phi_e=pir^2Q/(epsilon_0 pi R^2)=(Qr^2)/(epsilon_0R^2)
Now substitute these results into eq. (1) .
B*(2pi r)=mu_0epsilon_0 d/(dt)((Qr^2)/(epsilon_0R^2))
B*(2pi r)=(mu_0 r^2)/R^2 ((dQ)/(dt))
B=(mu_0 r)/(2 R^2 pi) ((dQ)/(dt))
B=(m_0 r)/(2 R^2 pi) I
B=(2*10^-7 (T*m)/A)(0.02 m)/(0.03 m)^2 (2.5 A)
B=1.11*10^-5 T