Question
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=3.00cm. The field magnitude is given by E=(4.50×10−3V/ms)t,
Where t is in seconds. What is the magnitude of the induced magnetic field at radial distances?
A. 2.00cm
B. 5.00cm
Solution
Since we have to find an induced magnetic field we will be using ampere’s circuital law which is ∫B.dl=μ0ε0dtdϕE
where ∫B.dlis the line integral of B around a closed path, μ0is the permeability of free space.
We will find the area of the circle by A=πr2 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 dtdE=4.50×10−3V/ms=0.0045V/ms
We will find the area of the circle by using the formula A=πr2
For r=2.00cm
r=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=21ε0μ0r(0.0045)
=21(8.85×10−12C2/Nm2)(4π×10−7Tm/A)(0.02m)(0.0045)
=5.01×10−22T
This is the magnitude of the induced magnetic field at a radial distance r=2.00cm
B. For r⟩R the above expression will change into
B(2πr)=μ0ε0πR2(0.0045)
Substituting the values of r=0.05m and R=0.03m we get,
B=4.51×10−22T
This is the magnitude of the induced magnetic field at a radial distance r=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
The product of current enclosed by the path and permeability of the medium equals the integral of magnetic field density (B) along an imaginary closed path.