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A sudden change in electric field will cause just magnetic field or changing magnetic field ?

Once the electric field is established and is not changing then what will happen to the magnetic field which was caused by the changing electric field?

Will the magnetic field (which was caused by the changing electric field) remains constant or will collapse ?

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A changing electric field causes a magnetic field and vice versa. This is given by the two Maxwell's equations.

$$\mathbf E\cdot d\mathbf l=-\dfrac{d\phi_B}{dt}$$

This is the well known Faraday's law and states that a changing magnetic field creates an electric field (or induces an EMF).

This induced electric field will be constant as long as the time derivative of magnetic flux is constant.

Now we have another equation which discusses the induction of a magnetic field with a changing electric field. This equation is also known as the Ampere-Maxwell equation.

$$\mathbf B\cdot d\mathbf l=\mu_0\epsilon_0\dfrac{d\phi_E}{dt}+\mu_0I$$

Let's assume the term $\mu_0I$ to be zero for a while because that only matters when there is a current moving through a wire, then we're just left with

$$\mathbf B\cdot d\mathbf l=\mu_0\epsilon_0\dfrac{d\phi_E}{dt}$$

which is the displacement current term and will be non-zero in case of a changing electric field. It simply ensures continuity of the magnetic field like in the case of a charging-discharging capacitor connected to an AC supply.

It works in the same manner as the previous one. That means, if the rate of change of electric field is constant, the magnetic field thus produced will also be constant.

Once the electric field is established and is not changing then what will happen to the magnetic field which was caused by the changing electric field?

We should notice that in both the cases, once the change stops i.e. the derivative becomes $0$ the left hand side of the equation also goes to $0$. So yes, if the change stops, the generation of the other field is also stopped.

Will the magnetic field (which was caused by the changing electric field) remain constant or collapse?

It will remain constant as long as $\dfrac{d\phi_E}{dt}$ is constant. But as you slow down and finally stop changing the electric field, the magnetic field will collapse as in the case of a DC capacitor circuit after a very long time when the current goes to $0$.

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  • $\begingroup$ "But as you slow down and finally stop changing the electric field, the magnetic field will collapse..." so it means that if the change in electric field suddenly stops then the magnetic field caused by the change is electric field will just collapse ? @user8718165 $\endgroup$ – Alex Dec 18 '19 at 18:46
  • $\begingroup$ "if electric field is constant, the magnetic field thus produced will also be constant. " so it means that if the change is not constant then the magnetic field will also be not constant and will be varying. right ? $\endgroup$ – Alex Dec 18 '19 at 18:48
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    $\begingroup$ Hi Alex, I forgot about about your question. Yes, if you abruptly stop changing the E-field the the B-field will just collapse. But in reality the change in derivative from "some number" to $0$ will be smooth. For eg. you can read about a DC RL circuit. $\endgroup$ – user8718165 Dec 27 '19 at 3:42
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    $\begingroup$ @Alex In response to your second question, yes, that's true. That way the E and B fields become self sustaining. That's how an EM wave is generated. $\endgroup$ – user8718165 Dec 27 '19 at 3:46
  • $\begingroup$ can you please answer this question? physics.stackexchange.com/questions/517093/… $\endgroup$ – Alex Dec 27 '19 at 6:49
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Assuming that there are no currents, the divergence and curl of the magnetic field satisfy Maxwell's equations: $$\boldsymbol{\nabla} \cdot \mathbf B(\mathbf x,t) = 0,$$ $$\boldsymbol{\nabla} \times \mathbf B(\mathbf x,t) = \frac{1}{c^2} \frac{\partial \mathbf{E}(\mathbf x,t)}{\partial t}.$$ Assuming that the time-dependence of the total electric field is known, the formal solution to this (along with the condition that $\mathbf E, \mathbf B \rightarrow 0$ at infinity) is $$\mathbf{B}(\mathbf x,t) = \frac{1}{4 \pi c^2} \int_{\mathbb R^3} d^3\mathbf x' \ \frac{\partial \mathbf{E}(\mathbf x',t)}{\partial t} \times \frac{\mathbf x - \mathbf x'}{|\mathbf x - \mathbf x'|^3}.$$ This can be derived by analogy with the Biot-Savart law, since the current density $\mathbf J$ is simply replaced by $\epsilon_0 \partial \mathbf E/\partial t$ in the curl equation. Do note however that $\mathbf E(\mathbf x,t)$ written here is the net electric field, which is not necessarily the applied field; the time-changing magnetic field itself creates an electric field that is included in $\mathbf E$ in the above expression.

You can see from this equation that in general, a time-dependent electric field can create a time-dependent magnetic field, which dies down after the electric field has settled on a constant value (i.e. $\partial \mathbf E/\partial t \sim 0$).

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