Question

32.2 induced magnetic fields
an electric field exists between a pair of circular metal plates measuring 3.00 m in radius. the field is uniform across the surface of the plates but increases in strength at a rate given by ( e(t)=at^{2} ), starting at ( t = 0 ) and persisting for 44.0 seconds. the constant, a, has a value of ( 45800vcdot m^{-1}cdot s^{-2} ). how strong will the magnetic field be on the edge of the plates at the end of the 44.0 second interval?
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Explanation:

Step1: Find the rate of change of electric field

Given \(E(t)=At^{2}\), the derivative \(\frac{dE}{dt} = 2At\).
Substitute \(A = 45800\space V\cdot m^{-1}\cdot s^{-2}\) and \(t = 44.0\space s\)
\(\frac{dE}{dt}=2\times45800\times44.0\)
\(\frac{dE}{dt}=4030400\space V\cdot m^{-1}\cdot s^{-1}\)

Step2: Use Ampere - Maxwell law

The Ampere - Maxwell law for a circular loop of radius \(r\) (edge of the plates) is \(\oint\vec{B}\cdot d\vec{l}=\mu_{0}\epsilon_{0}\frac{d\Phi_{E}}{dt}\)
For a circular loop, \(\oint\vec{B}\cdot d\vec{l}=B(2\pi r)\) and \(\Phi_{E}=EA\) (where \(A=\pi r^{2}\) is the area of the plate), so \(\frac{d\Phi_{E}}{dt}=\frac{d(EA)}{dt}=A\frac{dE}{dt}\) (since \(A=\pi r^{2}\) is constant)
\(B(2\pi r)=\mu_{0}\epsilon_{0}(\pi r^{2})\frac{dE}{dt}\)
Solve for \(B\): \(B=\frac{\mu_{0}\epsilon_{0}r}{2}\frac{dE}{dt}\)
We know that \(\mu_{0} = 4\pi\times 10^{-7}\space T\cdot m/A\) and \(\epsilon_{0}=8.85\times 10^{-12}\space C^{2}/(N\cdot m^{2})\), \(r = 3.00\space m\)
Substitute the values:

$$
LATEXBLOCK0
$$

$$
LATEXBLOCK1
$$

$$
LATEXBLOCK2
$$

$$
LATEXBLOCK3
$$

$$
LATEXBLOCK4
$$

$$
LATEXBLOCK5
$$

$$
LATEXBLOCK6
$$

$$
LATEXBLOCK7
$$

Answer:

\(6.72\times 10^{-10}\space T\)