Question

An electric field of 100 V/m is directed outward from the plane of a circular area with radius 4.0 cm. If the electric field increases at a rate of 10 V/ms, determine the magnitude of the magnetic field at a radial distance 10.0 cm away from the center of the circular area. 1.1 × 10-15 T 3.4 × 10-15 T 5.9 × 10-15 T 8.9 × 10-15 T 9.7 × 10-15 T

Answer 1

Answer:

The magnitude of the magnetic field is $8.9\times10^{-19}\ T$

Explanation:

Given that,

Electric field = 100 V/m

Radius = 4.0 cm

Electric field increase at a rate = 10 V/ms

Radial distance = 10.0 cm

We need to calculate the magnetic field

Using Gauss's law

$\oint{\vec{E}\cdot\vec{dA}}=\phi_{E}$

$\dfrac{dE}{dt}A=\dfrac{d\phi_{E}}{dt}$

$\dfrac{dE}{dt}(\pi r^2)=\dfrac{d\phi_{E}}{dt}$

We need to calculate the $\dfrac{d\phi}{dt}$

$\dfrac{d\phi}{dt}=10\times\pi\times(4.0\times10^{-2})^2$

$\dfrac{d\phi}{dt}=0.0503\ Nm^2/C.s$

According to Ampere Maxwell law

$\oint{\vec{B}\cdot \vec{ds}}=\mu_{0}(I+\epsilon_{0}\dfrac{d\phi_{E}}{dt})$

$\oint{\vec{B}\cdot\vec{ds}}=\mu_{0}I+\mu_{0}\epsilon_{0}\dfrac{d\phi_{E}}{dt})$

Electric field is zero inside the circle.

$\oint{\vec{B}\cdot \vec{ds}}=\mu_{0}\epsilon_{0}\dfrac{d\phi_{E}}{dt})$

$B(2\pi\times10.0\times10^{-2})=4\pi\times10^{-7}\times8.85\times10^{-12}\times0.0503$

$B=\dfrac{4\pi\times10^{-7}\times8.85\times10^{-12}\times0.0503}{2\pi\times10.0\times10^{-2}}$

$B=8.9\times10^{-19}\ T$

Hence, The magnitude of the magnetic field is $8.9\times10^{-19}\ T$