I"ve just began studying Maxwell"s equations today and what really had my attention is Ampere"s law, the second term in particular.$$intvec B cdot dvec l=μ_0I_encl+μ_0ε_0fracdΦ_Edt$$

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.

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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 ?


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Folshort
edited Jan 27 "17 at 8:17
John Katsantas
asked Jan 27 "17 at 8:06
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John KatsantasJohn Katsantas
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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


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answered Jan 27 "17 at 8:51
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Claudio Avi ChamiClaudio Avi Chami
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Wiki - displacement current: -

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Quotation: -

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

B = $dfracmu_0 I_D2pi r$.

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 $dfracdPhi_Edt$ is ∂E/∂t in the wikipedia quote.

See more: ¿ Mauricio Ochmann Y Su Esposa Maria Jose, Quiénes Han Sido Las Novias De Mauricio Ochmann

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


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Folshort
edited Jan 27 "17 at 9:07
answered Jan 27 "17 at 8:54
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Andy akaAndy aka
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Here is a diagram of a capacitor which is charging with and amperian loop shown in blue and the amperian surface shown in pink.

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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.

$$displaystyle oint_ m loop vec B cdot dvec l = mu_o I_ m surface+ mu_oepsilon_o dfrac dPhi_ m Edt$$

Left hand side$$displaystyle oint_ m loop vec B cdot dvec l = 2 pi r B$$

Right hand side$$mu_o I_ m surface = 0$$

For a parallel plate capacitor $$E = dfrac sigma epsilon_o$$ where $$sigma$$ is the surface charge density which is equal to $$dfracQpi R^2$$

$$Rightarrow E = dfracQepsilon_o pi R^2 Rightarrow Phi_ m E = dfracQepsilon_o pi R^2 pi r^2 = dfracQ r^2epsilon_o R^2$$

$$Rightarrow mu_oepsilon_o dfrac dPhi_ m Edt= dfracmu_o I r^2R^2$$ because $$dfracdQdt=I$$

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