Question

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.

Short answer

Answer: The magnitude of the magnetic field is approximately \(2.012 \times 10^{-8} \mathrm{T}\) and the direction is clockwise when looking at the circular planar surface from above.

Step-by-step solution

Understanding Faraday's law of electromagnetic induction

According to Faraday's law of electromagnetic induction, the change in electric flux through a surface induces a magnetic field around the surface: $$\oint \mathbf{B} \cdot \mathrm{d} \mathbf{A} = - \mu_{0} \epsilon_{0} \frac{\mathrm{d} \Phi_{\mathrm{E}}}{\mathrm{d}t}$$ In this problem, the electric field changes at a constant rate, so the rate of change of electric flux \(\frac{\mathrm{d} \Phi_{\mathrm{E}}}{\mathrm{d}t}\) will be constant as well.

Finding the change in the electric field

The rate of change in electric field is given as \(10.0 \mathrm{~V} /(\mathrm{m} \mathrm{s})\).

Calculate the change in electric flux

Since the electric field is perpendicular to the circular planar surface, the change in the electric flux can be determined as follows: $$\frac{\mathrm{d} \Phi_{\mathrm{E}}}{\mathrm{d}t} = \frac{\mathrm{d}(E \cdot A)}{\mathrm{d}t} = A \frac{\mathrm{d}E}{\mathrm{d}t}$$ where \(A\) is the area of the circular surface and \(\frac{\mathrm{d}E}{\mathrm{d}t}\) is the rate of change of the electric field. The area of a circle can be computed using the formula \(A = \pi r^{2}\). Here, \(r = 6.00 \mathrm{~cm} = 0.0600 \mathrm{~m}\). Therefore, $$A = \pi (0.0600 \mathrm{~m})^2 = 0.01131 \mathrm{~m^2}$$ Now, we can calculate the change in electric flux: $$\frac{\mathrm{d} \Phi_{\mathrm{E}}}{\mathrm{d}t} = 0.01131 \mathrm{~m^2} \cdot 10.0 \mathrm{~V} /(\mathrm{m} \mathrm{s}) = 1.131 \times 10^{-1} \mathrm{~V m/s}$$

Applying Faraday's law

Now, we can apply Faraday's law to find the magnetic field encircling the circular planar surface. With the given \(\frac{\mathrm{d} \Phi_{\mathrm{E}}}{\mathrm{d}t}\), we have: $$\oint \mathbf{B} \cdot \mathrm{d} \mathbf{A} = - \mu_{0} \epsilon_{0} (1.131 \times 10^{-1} \mathrm{~V m/s })$$

Determining the magnetic field at a specific radial distance

Let's consider a circle with radius \(r' = 0.100 \mathrm{~m}\) encircling the circular planar surface. Then, we will assume a uniform magnetic field along this circular path and integrate the magnetic field over the circumference to find the magnitude of the magnetic field. The magnetic field will have a tangencial direction to the circle and the direction can be determined by the right-hand-rule/screw rule. $$B \oint r' \mathrm{d}\theta = - \mu_{0} \epsilon_{0} (1.131 \times 10^{-1} \mathrm{~V m/s })$$ Dividing by the circumference \(2\pi r'\), we get the magnitude of the magnetic field, B: $$B = \frac{- \mu_{0} \epsilon_{0} (1.131 \times 10^{-1} \mathrm{~V m/s })}{2\pi (0.100 \mathrm{~m})}$$ Plug in the values of \(\mu_{0} = 4\pi \times 10^{-7} \mathrm{~T m/A}\) and \(\epsilon_{0} = 8.85 \times 10^{-12} \mathrm{~C^2/(N m^2)}\): $$B = \frac{- 4\pi \times 10^{-7} \mathrm {~T m/A} \cdot 8.85 \times 10^{-12} \mathrm{~C^2/(N m^2)} \cdot (1.131 \times 10^{-1} \mathrm{~V m/s })}{2\pi (0.100 \mathrm{~m})}$$ $$B = - 2.012 \times 10^{-8} \mathrm{~T}$$

Final answer

The magnitude of the magnetic field at a radial distance of 0.100 m (10.0 cm) away from the center of the circular area is approximately \(2.012 \times 10^{-8} \mathrm{~T}\). The negative sign indicates that the direction of the magnetic field is clockwise when looking at the circular planar surface from above.