
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 $E = (4.50 \times {10^{ - 3}}V/ms)t$,
Where $t$ is in seconds. What is the magnitude of the induced magnetic field at radial distances?
A. $2.00cm$
B. $5.00cm$
Answer
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Hint: Since we have to find an induced magnetic field we will be using ampere’s circuital law which is $\int {B.dl = {\mu _0}{\varepsilon _0}\dfrac{{d{\phi _E}}}{{dt}}} $
where $\int {B.dl} $is the line integral of B around a closed path, \[{\mu _0}\]is the permeability of free space.
We will find the area of the circle by $A = \pi {r^2}$ 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 $\dfrac{{dE}}{{dt}} = 4.50 \times {10^{ - 3}}V/ms = 0.0045V/ms$
We will find the area of the circle by using the formula $A = \pi {r^2}$
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.
$B = \dfrac{1}{2}{\varepsilon _0}{\mu _0}r(0.0045)$
$ = \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 \times {10^{ - 22}}T$
This is the magnitude of the induced magnetic field at a radial distance $r = 2.00cm$
B. For $r\rangle R$ the above expression will change into
$B(2\pi r) = {\mu _0}{\varepsilon _0}\pi {R^2}(0.0045)$
Substituting the values of $r = 0.05m$ and $R = 0.03m$ we get,
$B = 4.51 \times {10^{ - 22}}T$
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 $\overrightarrow B $
The product of current enclosed by the path and permeability of the medium equals the integral of magnetic field density ($\overrightarrow B $) along an imaginary closed path.
where $\int {B.dl} $is the line integral of B around a closed path, \[{\mu _0}\]is the permeability of free space.
We will find the area of the circle by $A = \pi {r^2}$ 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 $\dfrac{{dE}}{{dt}} = 4.50 \times {10^{ - 3}}V/ms = 0.0045V/ms$
We will find the area of the circle by using the formula $A = \pi {r^2}$
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.
$B = \dfrac{1}{2}{\varepsilon _0}{\mu _0}r(0.0045)$
$ = \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 \times {10^{ - 22}}T$
This is the magnitude of the induced magnetic field at a radial distance $r = 2.00cm$
B. For $r\rangle R$ the above expression will change into
$B(2\pi r) = {\mu _0}{\varepsilon _0}\pi {R^2}(0.0045)$
Substituting the values of $r = 0.05m$ and $R = 0.03m$ we get,
$B = 4.51 \times {10^{ - 22}}T$
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 $\overrightarrow B $
The product of current enclosed by the path and permeability of the medium equals the integral of magnetic field density ($\overrightarrow B $) along an imaginary closed path.
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