# What is the actual meaning of magnetic field and electric field in Electromagnetic waves?

I know that electromagnetic waves are composed of electric field and magnetic field but I don't understand what magnetic field is. Would you help me understand its physical meaning?

What is the difference between the magnetic field of electromagnetic field that can travel a very long distance and the magnetic filed of an ordinary magnet can travel few centimetres only. Why the magnetic field of a magnet can not travel a long distance like electromagnetic waves?

Thank you,

There isn't anything special about the magnetic field because your argument applies to an electric field as well. If I have a point charge then the electric field falls away as $1/r^2$, which is pretty quick. If I have an electric dipole then the electric field falls away as $1/r^3$, which is even quicker.

Incidentally, a bar magnet behaves approximately as a magnetic dipole and that's why its magnetic field falls away so fast - because it falls as $1/r^3$ just like the field from an electric dipole.

So the magnetic and electric fields behave the same way as far as their distance dependence is concerned, and indeed they behave the same way in an electromagnetic wave. Electric and magnetic fields need charges (moving charges for the magnetic field) to create them, and the distance from the charges affects their strength.

However the electromagnetic plane wave is a bit of a special case where the electric and magnetic fields combine to keep each other oscillating continuously without needing any charges.

The magnetic field in radiation that travels a long way is usually not due to magnetic dipoles such as the dipole moments of elementary particles. It is usually caused by the magnetic field from accelerating charges.

So let's talk about what the field at a point $\vec r$ at time $t$ is equal to due to a charge. You start at your point $\vec r$ and you go out to a sphere of radius $d$ and then ask yourself if there were any charges there at time $t-d/c$ (if there wasn't then do out to a different radius) and if there was that means you'd be able to see that charge at $\vec r$ at time $t$ so that's what you are looking for. If so let $\vec w$ be the position of that charge, let $q$ be the amount of charge, let $\vec a$ be the acceleration of the charge, and let $\vec v$ be the velocity of the charge and make sure all of them are evaluated at time $t-d/c$ then you have what you need.

Now we can compute the electric and magnetic fields due to that charge based on those three vectors. To compute the fields you can use $\vec d=\vec r -\vec w$ and $\vec u=c\hat d-\vec v$ and get (adapted from Griffiths' Introduction to electrodynamics):

$$\vec E(\vec r,t)=\frac{q}{4\pi\epsilon_0}\frac{d}{(\vec d\cdot \vec u)^3}\left[(c^2-v^2)\vec u+\vec d\times(\vec u\times \vec a)\right]$$

and $$\vec B(\vec r,t)=\frac{1}{c}\hat d\times \vec E(\vec r,t).$$

Recall that the $t$ is not the time when the charge had the position $\vec w,$ velocity $\vec v,$ and acceleration $\vec a.$ Those are at an earlier time, the time back when the charge broadcasted its position $\vec w,$ velocity $\vec v,$ and acceleration $\vec a$ just in time to arrive at $\vec r$ at time $t.$

Now, we see that it is term with the acceleration $\vec a$ that falls off the least, so while it might start off small, when you are super far away, that's the term that dominates.