How can the effect of electric and magnetic fields on the motion of a particle be distinguished? It's a conceptual question about magnetic field: Suppose a charged particle moves under the influence of an electric and a magnetic field, how do we distinguish the effect of the fields on the motion of the particle?
 A: We can very well distinguish between magnetic and electric fields by simply observing the movement of a charged particle.
As it was said correctly, the magnetic field does not do any work on the particle, meaning it does not accelerate it linearly. It does however apply a force that is perpendicular to the direction of motion. To illustrate, we assume a particle that moves in the x-direction in a homogenous magnetic field in the y-direction. As you surely know, the Lorentz force is given by $$\vec F = q \vec v \times \vec B$$, meaning that the force acts in the z-direction on the particle (which we for now assume to be positive). What will happen is that the charge will follow a circular path in that magnetic field.
A beautiful experiment is to use electrons in a Helmholtz coil pair (two large coils with a certain separation that creates a nearly homogenous magnetic field in the center):

Consider a homogenous electric field as a contrast: the force on the charge is given by $$\vec F = q \vec E$$ and is clearly directed in the direction of the electric field and not at right angles to it. In the simple case of $\vec v$ going in the same direction as $\vec E$, the charge will be accelerated. If it is at right angles, the charge will be deflected, just as you can imagine in the case of a horizontal throw in a homogenous gravitational field.

Depending on how you are allowed to prepare the experiment, it becomes fairly simple to see what effects the magnetic or electric fields have on a charge.
It must be noted, however, that electricity and magnetism are of course strongly interrelated, and the above discussion breaks down if you assume moving inertal frames. Special relativity connects those phenomena.
A: Let's assume we know the position $x(t)$ and the velocity $\dot x(t)$ of the particle on an interval of time $t$, and that we know the only force is the Lorentz force. Can we completely determine the electric and magnetic field at any point of the arc of trajectory?
Non-relativistic case
We will prove that at any point of the trajectory,


*

*we can determine the component of the electric field tangent to the trajectory;

*we cannot determine the component of the magnetic field tangent to the trajectory;

*knowing two of the field components perpendicular to the tangent to the trajectory allows to know the remaining two.


Let's introduce the arclength $s$, Frenet basis $(T, N, B)$, and the curvature $\rho$:
$$\begin{align}
\dot{x} &= \dot{s}T,\\
\ddot{x} &= \ddot{s}T+\rho\dot{s}^2N.
\end{align}$$
For the electric and magnetic field, I will use the subscripts $t$, $n$ and $b$ to denote their components in Frenet basis. Newton's law 
$$m\ddot x(t)=q(E+\dot x(t)\times B).$$
then reads
$$m(\ddot{s}T+\rho\dot{s}^2N)=q(e_tT+e_nN+e_bB+\dot{s}(b_nB-b_bN)),$$
i.e.
$$\begin{align}
e_t &= \frac{m}{q}\ddot{s},\\
e_n-\dot{s}b_b &= \frac{m}{q}\rho\dot{s}^2,\tag{1}\\
e_b+\dot{s}b_n &= 0.\tag{2}
\end{align}$$
QED.
Relativistic case
Let's prove the following:


*

*Same conclusions as for the non-relativistic case.


The right-hand side of Newton's law does not change while its left-hand side becomes $\dot{p}$ with $p=m\gamma\dot x$ but
$$\gamma = (1-\dot{s}^2)^{-1/2}.$$
Therefore
$$\dot{p} = m\gamma^3\ddot{s}\dot{x} + m\gamma\dot{x}.$$
Thus we can reuse the non-relativistic analysis by adding a term $m\gamma^3\ddot{s}\dot{s}T$ and replacing $m$ by $m\gamma$ everywhere else. Thus
$$\begin{align}
e_t &= \frac{m}{q}\gamma\ddot{s}(1+\gamma^2\dot{s}),\\
e_n-\dot{s}b_b &= \frac{m}{q}\gamma\rho\dot{s}^2,\tag{1}\\
e_b+\dot{s}b_n &= 0.\tag{2}
\end{align}$$
QED.
A: The magnetic field does no work on the particle and hence its energy remains constant. The effect of the electric field changes its energy.
