Magnetic field of a moving charged particle

[Crazymechanic]

Or maybe your just not getting the answer provided to you.

There is no magnetic field of a moving particle in it's own frame of reference if I even can call it so.
The magnetic field is only to a observer which moves at a different speed than that of the moving particle or is stationary with respect to it.

there is no difference from a static electric field to that of a particle because the field of a single charged particle is fixed in value and is static if that is what you wanted to know.

 

[Dale]

there is no difference from a static electric field to that of a particle because the field of a single charged particle is fixed in value and is static if that is what you wanted to know.

There is a difference between the electric field of a static charge and a moving charge. Again, I point to the Lienard Wiechert fields:
http://en.wikipedia.org/wiki/Liénar...onding_values_of_electric_and_magnetic_fields

In these equations the velocity as a fraction of the speed of light is given by $\beta$. Thus, the case of a non-accelerating charge is given by $\dot{\beta}=0$ while the case of a stationary particle is given by $\beta=0$. Even if you don't follow the meaning of the equation you can clearly see that it is a function of $\beta$ which means that the field depends on the velocity.

 

[WannabeNewton]

there is no difference from a static electric field to that of a particle because the field of a single charged particle is fixed in value and is static if that is what you wanted to know.

Without even looking at the solution, this clearly cannot be true. Faraday's law and Gauss's law read $\vec{\nabla} \cdot \vec{E} = 4\pi \rho$ and $\vec{\nabla} \times \vec{E} = -\frac{1}{c}\partial_t \vec{B}$ so clearly the solution for an arbitrarily accelerating charge (time-varying delta function source) is in general different from the solution for the special case of a static charge, which simply has $\vec{\nabla} \cdot \vec{E} = 4\pi Q\delta^3 (\vec{r} - \vec{r}')$ and $\vec{\nabla} \times \vec{E} = 0$ where $\vec{r}'$ fixed.

 

[Crazymechanic]

sorry wrong button .:)