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Question: In the figure, a uniform electric field is directed out of the page within a circular region of radi...

In the figure, a uniform electric field is directed out of the page within a circular region of radius R=3.00cmR = 3.00cm. The field magnitude is given by E=(4.50×10−3V/ms)tE = (4.50 \times {10^{ - 3}}V/ms)t,
Where tt is in seconds. What is the magnitude of the induced magnetic field at radial distances?
A. 2.00cm2.00cm
B. 5.00cm5.00cm

Explanation

Solution

Since we have to find an induced magnetic field we will be using ampere’s circuital law which is ∫B.dl=μ0ε0dϕEdt\int {B.dl = {\mu _0}{\varepsilon _0}\dfrac{{d{\phi _E}}}{{dt}}}
where ∫B.dl\int {B.dl} is the line integral of B around a closed path, μ0{\mu _0}is the permeability of free space.
We will find the area of the circle by A=πr2A = \pi {r^2} and put the value of radial distance in the above formula. We will get the magnitude of the induced magnetic field.

Complete step by step solution:
A. We have been given that dEdt=4.50×10−3V/ms=0.0045V/ms\dfrac{{dE}}{{dt}} = 4.50 \times {10^{ - 3}}V/ms = 0.0045V/ms
We will find the area of the circle by using the formula A=πr2A = \pi {r^2}
For r=2.00cmr = 2.00cm
r=0.02mr = 0.02m
Using ampere’s circuital law, Ampere's circuital law states that the integrated magnetic field around a closed loop to the electric current passing through the loop.
B=12ε0μ0r(0.0045)B = \dfrac{1}{2}{\varepsilon _0}{\mu _0}r(0.0045)
=12(8.85×10−12C2/Nm2)(4π×10−7Tm/A)(0.02m)(0.0045)= \dfrac{1}{2}(8.85 \times {10^{ - 12}}{C^2}/N{m^2})(4\pi \times {10^{ - 7}}Tm/A)(0.02m)(0.0045)
=5.01×10−22T= 5.01 \times {10^{ - 22}}T
This is the magnitude of the induced magnetic field at a radial distance r=2.00cmr = 2.00cm
B. For r⟩Rr\rangle R the above expression will change into
B(2πr)=μ0ε0πR2(0.0045)B(2\pi r) = {\mu _0}{\varepsilon _0}\pi {R^2}(0.0045)
Substituting the values of r=0.05mr = 0.05m and R=0.03mR = 0.03m we get,
B=4.51×10−22TB = 4.51 \times {10^{ - 22}}T
This is the magnitude of the induced magnetic field at a radial distance r=5.00cmr = 5.00cm

Note:
Ampere's law is one of the Maxwell's equations, which is useful when the structure involved is symmetric, to evaluate the line integral of B→\overrightarrow B
The product of current enclosed by the path and permeability of the medium equals the integral of magnetic field density (B→\overrightarrow B ) along an imaginary closed path.