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I have studied in Introduction to electrodynamics (Griffiths) that magnetic field is actually due to effects of relativity

unequal Lorentz contraction of the positive charge and negative lines, a current- carrying wire that is electrically neutral in one inertial system will be charged in another.

(this was ,'as explained in Griffiths.)

But here (MIT website, chapter 9, http://ocw.mit.edu/courses/physics/8-02t-electricity-and-magnetism-spring-2005/lecture-notes/ ) I have seen,

a single charge also produces magnetic field

but how this is possible? How can I imagine contraction of single charge (not a line of charge).

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    $\begingroup$ Don't mistake a contrived situation in which it is easy to calculate an effect for the cause or definition of the effect. Griffiths sets up the former. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Nov 9, 2014 at 18:32

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According to the Lorentz transformations of the electric and magnetic fields, there is also a magnetic field in a frame of reference moving relative to another frame of reference in which there is just a static electric field.

So, assuming an isolated point charge, in the frame of reference in which the charge is at rest, there is only a static, radially directed electric field.

But, in a relatively moving frame of reference, there is also a magnetic field.

Put another way, it isn't that we apply the Lorentz transformation to the electron and then calculate the field, we apply the Lorentz transformation to the electromagnetic field over all space.

The case of a neutral wire with steady current is a special case with a symmetry that allows us to derive some results without directly applying the Lorentz transformation to the fields themselves.

In summary, to understand the answer to your question fully, you must appreciate that the electric and magnetic fields themselves transform according to the Lorentz transformations - relatively moving observers 'see' different electric and magnetic fields.

Update:

Since you are reading Griffiths' "Introduction to Electrodynamics", please see Example 12.15 (of the 4th edition) Magnetic field of a point charge in uniform motion where Griffiths derives the magnetic field by transforming from the static electric field of the point charge's rest frame into a relatively moving frame.

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Related How do moving charges produce magnetic fields?

A moving charge $q$ with velocity $\vec{v}$ produces a current

$$\vec{j}=\rho(r) \vec{v}$$

($\rho$ charge density associated to charge $q$, $\vec{j}$ current density, ref)

and by Maxwell's equation for the magnetic field (Ampere Law):

$$\nabla \times \vec{B} = \mu_0\vec{J} + \partial \vec{E}/\partial t$$

Magnetic field is produced.

In special relativity (inertial frames), in a moving frame of the charged partile, the magnetic field calculated above can also be seen as the relativistic correction to the electrostatic field of the rest frame.

A Lorentz transformation makes this explicit.

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  • $\begingroup$ Nikos, your first equation is puzzling to me. $\endgroup$
    – Alfred Centauri
    Commented Nov 9, 2014 at 17:17
  • $\begingroup$ @AlfredCentauri, yes it is not exact actually (i'll have to look it up, dont remember exact form), but the equation states how a current is generated (or equivalent) to a moving charge $\endgroup$
    – Nikos M.
    Commented Nov 9, 2014 at 17:25
  • $\begingroup$ @AlfredCentauri, for example take a look at this lecture, eq for B under current density and equation for B under Magnetic Field of a Moving Charge (current density is $\vec{j}=q\vec{v}$) it is current density actually (that is why slightly differnt symbol) $\endgroup$
    – Nikos M.
    Commented Nov 9, 2014 at 17:28
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    $\begingroup$ Nikos, yes $\vec j$ is current density which is why I'm puzzled. Current density is in ampere's per square meter but $q\vec v$ has unit ampere meter. I suspect you meant to write something like $\vec j = \rho(\mathbf r) \vec v$ where the charge density is a delta function for a point charge. $\endgroup$
    – Alfred Centauri
    Commented Nov 9, 2014 at 17:35
  • $\begingroup$ @AlfredCentauri, yes it is an overloading of notation (where $Q$ means charge density), fixing that! $\endgroup$
    – Nikos M.
    Commented Nov 9, 2014 at 17:37
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A single charge cannot produce a static magnetic field. It can produce a time-dependent magnetic field. It is also a relativity effect because both electric and magnetic fields are changed with change of the reference frame.

In a moving reference frame (or when the charge moves) one observes not only electric, but also a magnetic field $\mathbf{H}'\propto\mathbf{V}\times\mathbf{E}$

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