An electric field of magnitude \(200.0 \mathrm{~V} / \mathrm{m}\) is directed perpendicular to a circular planar surface with radius \(6.00 \mathrm{~cm}\). If the electric field increases at a rate of \(10.0 \mathrm{~V} /(\mathrm{m} \mathrm{s}),\) determine the magnitude and the direction of the magnetic field at a radial distance \(10.0 \mathrm{~cm}\) away from the center of the circular area.

First, we need to calculate the induced electromotive force (EMF) in the circular planar surface due to the change in the electric field. According to Faraday's law of electromagnetic induction: \[\oint _C \mathbf{E} \cdot d\bold{ \ell} = -\frac {d\Phi _B}{dt}\] Where \(\mathbf{E}\) is the electric field, \(d\bold{ \ell}\) is the infinitesimal length element of a closed curve \(C\), and \(\frac {d\Phi _B}{dt}\) is the time derivative of the magnetic flux \(\Phi _B\). In this case, since the electric field is uniform and perpendicular to the circular area, the magnetic flux is: \[\Phi_B = E\pi r_e^2\] Where \(r_e = 6.00\mathrm{~cm} = 0.06\mathrm{~m}\) The time derivative of the magnetic flux is given by: \[\frac {d\Phi _B}{dt} = \frac {d(E\pi r_e^2)}{dt} = \pi r_e^2 \frac {dE}{dt}\] Then, we can substitute with the given values. \[\frac{d\Phi_B}{dt} =\pi (0.06\mathrm{~m})^2 (10.0 \mathrm{~V}/(\mathrm{m}\mathrm{s}))\]

Now, we need to find the induced current in the circular surface. To do that, we can rewrite Faraday's law as follows: \[I \oint _C \mathbf{B} \cdot d\bold{ \ell} = -\frac {d\Phi _B}{dt}\] Where \(I\) is the induced current and \(\mathbf{B}\) is the magnetic field. Since the direction of the induced current is tangential to the surface and perpendicular to the magnetic field, the expression for the induced current becomes: \[I = -\frac {1}{2\pi r_c}\frac {d\Phi _B}{dt}\] Where \(r_c = 10.0\mathrm{~cm} = 0.10\mathrm{~m}\) is the radial distance away from the center of the circular area.

Now, we will use the Biot-Savart law to find the magnetic field induced by the induced current at radial distance \(r_c\). The Biot-Savart law is given as follows: \[\mathbf{B} = \frac{\mu_0}{4\pi}\int \frac{I{}d\mathbf{l} \times \mathbf{r'}}{r'^3}\] For this circular surface, the induced magnetic field is: \[\mathbf{B} = \frac{\mu_0 I}{2r_c}\] Where \(\mu_0\) is the permeability of free space. Now, we can substitute the induced current and the given values: \[\mathbf{B} = \frac{(4\pi \times 10^{-7}\mathrm{Tm/A})}{2\cdot 0.10\mathrm{~m}}\cdot \left(-\frac {1}{2\pi 0.10\mathrm{~m}}\right)\frac {d\Phi _B}{dt}\]

Now, we can find the magnitude of the induced magnetic field, with direction given by the right-hand rule. Using the given values and the result from the previous steps: \[\mathbf{B} = 7.96 \times 10^{-8} \mathrm{T}\] The induced magnetic field is upwards if we take into account the right-hand rule and the direction of the electric field. Final answer, Magnitude: \(7.96 \times 10^{-8} \mathrm{T}\) Direction: Upwards