When a capacitor is charged, the electric field E, and hence the electric flux Φ, between the plates changes. This change in flux induces a magnetic field, according to Ampère's law as extended by Maxwell: ∮B⃗ ⋅dl⃗ =μ0(I+ϵ0dΦdt). You will calculate this magnetic field in the space between capacitor plates, where the electric flux changes but the conduction current I is zero. A parallel-plate capacitor of capacitance C with circular plates is charged by a constant current I. The radius a of the plates is much larger than the distance d between them, so fringing effects are negligible. Calculate B(r), the magnitude of the magnetic field inside the capacitor as a function of distance from the axis joining the center points of the circular plates. Express your answer in terms of μ0 and given quantities. B(r) =

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The magnitude of the magnetic field B(r) inside the capacitor as a function of distance from the axis is [tex]\mathbf{B(r) =\dfrac{\mu_o I r}{2 \pi a^2} }[/tex]

A charged capacitor usually cause changes between the plates of electric field E and the electric flux Φ

As seen in Ampere's law which is extended by Maxwell:

[tex]\mathbf{\oint B^{\to} .dl^{\to} = \mu_o \Big( I+\dfrac{d \phi }{dt} \Big)}[/tex]

To meet the equilibrium equation of electric charge, Maxwell modified Ampere's law by incorporating the displacement current through into electric current component.

The displacement current can be expressed by the relation;

[tex]\mathbf{I = \varepsilon_o \dfrac{dE}{dt} }[/tex]

From the given information, the displacement current density through area A can now be expressed as:

[tex]\mathbf{I = \varepsilon_o A \dfrac{dE}{dt} }[/tex]

Replacing the value of displacement current density into Maxwell modified Ampere Law, we have:

[tex]\mathbf{\oint B .dl= \mu_o I_{enclosed} }[/tex]

[tex]\mathbf{B \times (2 \pi r) = \mu_o I\Big (\dfrac{\pi r^2}{\pi a^2} \Big) }[/tex]

[tex]\mathbf{B(r) =\dfrac{\mu_o I r}{2 \pi a^2} }[/tex]

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