Respuesta :
(a) 0
The magnetic field strength insidea a parallel-plate capacitor with changing electric field can be found by applying Ampere's law:
[tex](2\pi r) B = \mu_0 I_D[/tex] (1)
where
[tex](2\pi r)[/tex] is the circumference of the circular line of radius r with axis coincident to the axis of the capacitor, used to calculate the magnetic field
B is the strength of the magnetic field
[tex]I_D[/tex] is the displacement current enclosed by the area of the circular line mentioned above, and it is equal to
[tex]I_D = \epsilon_0 \frac{d\Phi_E}{dt} = \epsilon_0 (\pi r^2) \frac{dE}{dt}[/tex] (2)
where
[tex]\frac{d\Phi_E}{dt}[/tex] is the rate of change of electric flux through the area enclosed by the line
[tex]\frac{dE}{dt}=1.0\cdot 10^6 V/m[/tex] is the rate of change of the electric field
Rewriting eq.(1), we find
[tex]B = \frac{\mu_0 \epsilon_0 r}{2}\frac{dE}{dt}[/tex]
which is valid for r < R (where R=5.0 cm is the radius of the plates of the capacitor).
In this part of the problem,
r = 0
since we are on the axis; so substituting r=0 inside the formula above, we find
B(0) = 0
(b) [tex]1.67\cdot 10^{-13}T[/tex]
In this part, we have
r = 3.0 cm = 0.03 m
The formula used in part (a) is still valid since r<R, so we can directly use it to find the magnitude of the magnetic field:
[tex]B = \frac{\mu_0 \epsilon_0 r}{2}\frac{dE}{dt}=\frac{(4\pi\cdot 10^{-7}H/m)(8.85\cdot 10^{-12}F/m)(0.03 m)}{2}(1.0\cdot 10^6 V/m)=1.67\cdot 10^{-13}T[/tex]
(c) [tex]1.98\cdot 10^{-13} T[/tex]
In this part, we have
r = 7.0 cm = 0.07 m
so here
r > R
therefore we need to substitute [tex](\pi r^2)[/tex] with [tex](\pi R^2)[/tex] in eq. (2), since the area through which the flux is calculated is only [tex](\pi R^2)[/tex] (there is no electric field outside the area of the capacitor). So we find
[tex]I_D = \epsilon_0 (\pi R^2) \frac{dE}{dt}[/tex]
and therefore
[tex]B = \frac{\mu_0 \epsilon_0 R^2}{2r}\frac{dE}{dt}=\frac{(4\pi\cdot 10^{-7}H/m)(8.85\cdot 10^{-12}F/m)(0.05 m)^2}{2(0.07 m)}(1.0\cdot 10^6 V/m)=1.98\cdot 10^{-13} T[/tex]
The magnitude of the magnetic field 3 cm from the axis is [tex]1.67\times 10^{-13} \rm \ T[/tex].
What is a magnetic field?
It is a vector field in which ferromagnetic objects and moving charges experience an influence.
The magnitude of the magnetic field can be calculated by the formula,
[tex]B =\dfrac { \mu _0 \epsilon_0 r} 2\times \dfrac {dE}{dt}[/tex]
Where,
μ - magnetic permeability = [tex]4\pi \times 10^{-7}{\rm \ H/m}[/tex]
r - distance = 3 cm
[tex]\dfrac {dE}{dt}[/tex] - rate of electic field = [tex]1.0 \times 10^6 \rm v/m s.[/tex]
Put the values in the formula,
[tex]B =\dfrac { 4\pi \times 10^{-7}{\rm \ H/m} (8.85\times 10^{-12}) (0.03)} 2\times(1.0\times 10^6)\\\\B = 1.67\times 10^{-13} \rm \ T[/tex]
Therefore, the magnitude of the magnetic field 3 cm from the axis is [tex]1.67\times 10^{-13} \rm \ T[/tex].
Learn more about the magnetic field:
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