Is there a magnetic field between capacitor plates while the capacitor is charging?

Question

I've just began studying Maxwell's equations today and what really had my attention is Ampere's law, the second term in particular. $$\int\vec B \cdot d\vec l=μ_0I_{encl}+μ_0ε_0\frac{dΦ_E}{dt}$$

Does this mean that a changing electric field can cause a magnetic field? For example, during the charging of a capacitor, between the plates where the electric field is changing.

I saw an exercise example where we changed the voltage across a capacitor and thus created a magnetic field between them.But some websites state that as long as there is no current - charge movement at the place of interest, there is no magnetic field being created. I read the same about the capacitor in particular. Could the example be wrong or is there a difference ?

Answer 1

The reason for the introduction of the 'displacement current' was exactly to solve cases like that of a capacitor. A magnetic field cannot have discontinuities, unlike the electric field (there are electric charges, but there are not magnetic monopoles, at least as far as we know in the Universe in its current state). There cannot be a magnetic field outside the capacitor and nothing inside. en.wikipedia.org/wiki/Displacement_current

Answer 2

Wiki - displacement current: -

Quotation: -

However, applying this law to surface S2, which is bounded by exactly the same curve ∂ S, but lies between the plates, provides:

B = \$\dfrac{\mu_0 I_D}{2\pi r}\$.

Any surface that intersects the wire has current I passing through it so Ampère's law gives the correct magnetic field. Also, any surface bounded by the same loop but passing between the capacitor's plates has no charge transport flowing through it, but the ε\$_0\$ ∂E/∂t term provides a second source for the magnetic field besides charge conduction current. Because the current is increasing the charge on the capacitor's plates, the electric field between the plates is increasing, and the rate of change of electric field gives the correct value for the field B found above.

Note that in the question above \$\dfrac{d\Phi_E}{dt}\$ is ∂E/∂t in the wikipedia quote.

The whole basis for electromagnetic wave propagation relies on displacement currrent producing a magnetic field.

Answer 3

Here is a diagram of a capacitor which is charging with and amperian loop shown in blue and the amperian surface shown in pink.

The area vector is in the same direction as the electric field $$\vec E$$ and so the positive direction around the loop is anticlockwise looking from the top - blue arrow.

$$\displaystyle \oint_{\rm loop} \vec B \cdot d\vec l = \mu_o I_{\rm surface}+ \mu_o\epsilon_o \dfrac {d\Phi_{\rm E}}{dt}$$

Left hand side
$$\displaystyle \oint_{\rm loop} \vec B \cdot d\vec l = 2 \pi r B$$

Right hand side
$$\mu_o I_{\rm surface} = 0$$

For a parallel plate capacitor $$E = \dfrac \sigma \epsilon_o$$ where $$\sigma$$ is the surface charge density which is equal to $$\dfrac{Q}{\pi R^2}$$

$$\Rightarrow E = \dfrac{Q}{\epsilon_o \pi R^2} \Rightarrow \Phi_{\rm E} = \dfrac{Q}{\epsilon_o \pi R^2} \pi r^2 = \dfrac{Q r^2}{\epsilon_o R^2}$$

$$\Rightarrow \mu_o\epsilon_o \dfrac {d\Phi_{\rm E}}{dt}= \dfrac{\mu_o I r^2}{R^2}$$ because $$\dfrac{dQ}{dt}=I$$

Equating the left hand side and the right hand side gives a value for the magnetic field at a distance r from the central axis of the capacitor

$$B = \dfrac{\mu_oIr}{2\pi R^2}$$ for $$0\le r\le R$$

and with r=R this gives the familiar $$B = \dfrac{\mu_oI}{2\pi R}$$