Answer
$6.40\times10^{-8}$ T, 0T.
Work Step by Step
Because charge is conserved, we know that the current in the wires is the same as the rate at which charge accumulates on the plates. This is the displacement current in this situation. Outside the plates of the capacitor, the displacement current plays the role of an actual current.
We are at a distance greater than the radius of the plates, so we may use the expression for the magnetic field of a long wire, equation 20–6.
$$B=\frac{\mu_oI}{2\pi R}=\frac{(4\pi \times10^{-7}T\cdot m/A)(32\times10^{-3}C/s)}{2 \pi (0.100m)}=6.40\times10^{-8}T$$
After the capacitor is fully charged, there will be no current. The plates are fully charged, so there will be no change in electric field or electric flux, so, as expected there will also be no displacement current. The magnetic field will be zero.
