I was taught that , in a electromagnetic radiation , changing electric field is produced due to oscillation of charge this changing electric field give rise to 'changing' magnetic field and this changing magnetic field again gives rise to changing electric field . And this process is continue to occur

But my question is : a changing electric field give rise to constant magnetic field it doesn't produced changing magnetic field

So , how all of these will be correct?

If there is a doubt on the behaviour of the magnetic field in electromagnetic radiation, the best is to check it out with Maxwell's free equations:

$$\nabla \times\mathbf{E} =-\frac{1}{c} \frac{\partial \mathbf{B}}{\partial t}$$

$$\nabla \times \mathbf{B} = \frac{1}{c} \frac{\partial \mathbf{E}}{\partial t}$$

$$\nabla\cdot \mathbf{E} =0\quad \text{and} \quad \nabla\cdot \mathbf{B}=0$$

$$\mathbf{E} = \mathbf{E}_0 e^{i(\mathbf{k}\cdot \mathbf{r} - \omega t)}$$

If we plug that in the first equation we get:

$$i \mathbf{k}\times \mathbf{E}_0 e^{i(\mathbf{k}\cdot \mathbf{r} - \omega t)} = -\frac{1}{c} \frac{\partial \mathbf{B}}{\partial t}$$

We could now integrate over $$t$$ in order to get the result for $$\mathbf{B}$$. But we just guess the form of $$\mathbf{B}$$:

$$\mathbf{B} = \mathbf{B}_0 e^{i(\mathbf{k}\cdot \mathbf{r} - \omega t)}$$

Upon plugging that in the precedent equation we get:

$$i \mathbf{k}\times \mathbf{E}_0 e^{i(\mathbf{k}\cdot \mathbf{r} - \omega t)} = i\frac{\omega}{c} \mathbf{B}_0 e^{i(\mathbf{k}\cdot \mathbf{r} - \omega t)} \tag{1}$$

We learn the following from this result:

We have dispersion equation $$|\mathbf{k}| = \frac{\omega}{c}$$

The magnetic field vector is orthogonal to the (direction of the) wave vector and the electric field vector. Using $$\nabla \cdot \mathbf{E} =0$$ we also find that the wave vector $$\mathbf{k}$$ is orthogonal to the electric field vector.

$$\mathbf{k}\cdot \mathbf{E} =0$$

We can check our guess by evaluating the second Maxwell equation:

$$\nabla \times \mathbf{B} = -\frac{\omega}{c} \mathbf{E}_0 e^{i(\mathbf{k}\cdot \mathbf{r} - \omega t)}$$

and find that our guess $$\mathbf{B} = \mathbf{B}_0 e^{i(\mathbf{k}\cdot \mathbf{r} - \omega t)}$$

also fulfills the second Maxwell equation by using the dispersion relation $$|\mathbf{k}|= \frac{\omega}{c}$$:

$$i\mathbf{k}\times \mathbf{B}_0 e^{i(\mathbf{k}\cdot \mathbf{r} - \omega t)} = - i\frac{\omega}{c} \mathbf{E}_0 e^{i(\mathbf{k}\cdot \mathbf{r} - \omega t)} \tag{2}$$

We carry out a last check: We turn out second result (2) into the first one (1). For this we vector multiply the last equation with $$\mathbf{k}$$ and swap rhs with lhs:

$$-\mathbf{k}\times \mathbf{E_0} \frac{\omega}{c} = \mathbf{k}\times (\mathbf{k} \times \mathbf{B}_0) = (\mathbf{k}\cdot \mathbf{B}_0)\mathbf{k} - k^2 \mathbf{B}_0 = -k^2\mathbf{B}_0$$

Because of $$\nabla\cdot \mathbf{B}=0$$ we know that $$\mathbf{k}\cdot \mathbf{B}_0 =0$$. Furthermore we know that $$k^2 =(\frac{\omega}{c})^2$$

So cancelling out a $$-\frac{\omega}{c}$$ we get our first result (1).

So we find that the the time and position dependence of the magnetic field is the same as for the electrical field, but the amplitude of both is the same in absolute values ( $$|\mathbf{E}_0| =|\mathbf{B}_0|$$. Note that the calculation was done in cgs-units where magnetic and electric field have the same units).

One can make the proof even more general in considering the fields (used here as check for their behaviour) as Fourier components of a more general behaviour on time and position. But this does not change the general picture, a varying electric field drives a varying magnetic field which again drives a varying electric field.

• Imagine a simple d.c circuit with closed key , the current is flowing through the wire because of changing electric field , due to chnaging E field , there generates a constant magnetic field. Is it any special case?
– user324098
Commented Jan 1, 2022 at 4:17
• Your post is about electromagnetic radiation. In that case the free Maxwell equations apply. If there are sources like a current, the Maxwell equations with sources apply. BTW: what do you mean with changing? Change in position or change in time ? Commented Jan 1, 2022 at 7:19
• Change in time. What is meant by Maxwell equation with sources
– user324098
Commented Jan 1, 2022 at 12:50
• When one has charge or a current this is considered as a source. However the electromagnetic radiation is a consequence of the sourceless Maxwell equations. If one has a wire with a D.C. current running, this creates a magnetic field which is time constant according to Ampere's law. No addtional electric field is necessary for this. Commented Jan 1, 2022 at 14:49

a changing electric field give rise to constant magnetic field

Only if derivative of the E-field is constant, what is a very special case, for a linear increasing or decreasing field. In general the derivative itself changes with time, what generates a changing magnetic field.

• Means , we have to consider double derivate of E field?
– user324098
Commented Jan 1, 2022 at 4:16

Not all of this is correct. This is correct: "changing electric field give rise to 'changing' magnetic field". And this is wrong according to Maxwell's equations. "a changing electric field give rise to constant magnetic field".

• Why , changing electric field do give rise to constant magnetic field
– user324098
Commented Dec 31, 2021 at 17:40
• Where did you get this information? Commented Dec 31, 2021 at 19:35
• Imagine a simple d.c circuit with closed key , the current is flowing through the wire because of changing electric field , due to chnaging E field , there generates a constant magnetic field
– user324098
Commented Jan 1, 2022 at 4:11
• @VaibhavTiwari what if I gave you the magnetic field $B_0\sin{(\omega t)}$, for some frequency $\omega$, what would the resulting electric field be? Commented Jan 1, 2022 at 19:21