Question

a circular surface lies perpendicular to an electric field which is uniform across the surface but starting at t = 0, the field increases in strength at a rate proportional to the square of the elapsed time. during this same time interval, the induced magnetic field within the surface

will decrease at a constant rate.
will increase with time as t².
will be constant in time.
will increase at a constant rate.

Explanation:

Step1: Recall Maxwell's equations

From Maxwell's equations, the displacement current \(I_d=\epsilon_0\frac{d\Phi_E}{dt}\). Given \(E\propto t^2\), \(\Phi_E = EA\) (where \(A\) is the area of the circular surface, constant). So \(\frac{d\Phi_E}{dt}\propto t^2\)'s derivative.

Step2: Calculate the derivative of \(\Phi_E\)

Differentiating \(\Phi_E\) with respect to \(t\), \(\frac{d\Phi_E}{dt}=A\frac{dE}{dt}\). Since \(E = kt^2\) (where \(k\) is a constant), \(\frac{dE}{dt}=2kt\). Then \(I_d=\epsilon_0A\frac{dE}{dt}\propto t\).

Step3: Use Ampere - Maxwell law

The Ampere - Maxwell law is \(\oint\vec{B}\cdot d\vec{l}=\mu_0(I_{enc}+I_d)\). For a circular loop of radius \(r\) (assuming the surface is a circle of radius \(r\)), \(\oint\vec{B}\cdot d\vec{l}=B(2\pi r)\) and \(I_{enc} = 0\) (no conduction current). So \(B=\frac{\mu_0I_d}{2\pi r}\). Since \(I_d\propto t\), \(B\propto t\).

Answer:

will increase at a constant rate.