Maxwell Equations and magnetic field?

Question

One of the maxwell equations state that the magnetic field induced around a closed loop is proportional to the electric current plus displacement current (rate of change of electric field) it encloses.

It means that either current or change in electric field or both can cause a magnetic field.

My question is that if both current and change in electric field cause magnetic field
then the magnetic field produced by current will add up into the magnetic field caused by change in electric field or both will cancel out eachother or they will combine to give anything different. Please avoid mathematical explaination. Just tell me in simple words the answer of my question.

Answer 1

If I understand your question correctly, you're asking whether the magnetic field induced by a current $\vec{J}$ and the magnetic field induced by a changing electric field $\partial \vec{E}/\partial t$ will reinforce each other, cancel out, or combine in some other weird way.

The answer is that there's no general rule. It depends on the relative directions of the current $\vec{J}$ and the $\partial \vec{E}/\partial t$, and one could easily think of situations where any one of these three options (reinforce, cancel, weird combination) occur.

It's like asking, without any context, whether two vectors $\vec{A}$ and $\vec{B}$ point in the same direction. it's a valid question, but you have to know something about the vectors in question before you can answer it.

Answer 2

It means that either current or change in electric field or both can cause a magnetic field.

It means that either current or change in electric field or both can cause a magnetic field.

The problem with this interpretation is that it is not clear how to determine the contribution due to electric displacement in practice. If changing electric field is present there is also some changing electric current.

Mathematically we could attempt to solve the Maxwell equation where the electric current is written off and only electric displacement current is used. It turns out then in most common situations, the magnetic field solving such artificial equation is effectively zero. This is exact result whenever electric field everywhere is given by gradient of a potential.

In case of an inductor with time-varying current, the electric field is not everywhere given accurately by a potential, so the contribution of displacement current to the magnetic field could be non-zero. However, the electric field is strong only near and inside the conductor. Inside the conductor, the displacement current has not always the same direction as the electric current, but for low frequency currents, it always has much lower magnitude. When determining the magnetic field, it can be usually ignored. The exception is the rapidly oscillating electric and magnetic fields of a radiating antenna.

For sinusoidal current of frequency $\omega$, the electric current density obeys $$ j =\sigma E $$ and electric displacement current density obeys $$ j_D=\epsilon_0 \omega E $$ For low frequency $\omega$, the displacement current is much less than electric current density and can be ignored. It begins to be comparable for frequencies around and higher than

$$ \omega = \frac{\sigma}{\epsilon_0} $$

Conductivity $\sigma$ depends on frequency (decreases for high frequencies) but if we simplify and assume the fiction that the conductor has conductivity of copper for direct current $\sigma = 6\text{E}7 $ A/(Vm), then the above frequency is of the order of $7\text{E}18$ Hz, which is in the range of X-rays (very fast oscillations). In reality conductivity gets lower with increasing frequency so this will happen sooner, for lower frequencies. I'd guess for frequencies where radiation emissions begin to be substantial, say MHz-GHz.