Charge launched parallel to magnetic field lines
If we launch an electric charge with speed v into a uniform magnetic field B , this charge will make a uniform movement within the magnetic field. Therefore, the different types of trajectory assumed by the charge depend on the angle at which it was thrown into the magnetic field.
In the first situation analyzed, we will assume that the charge qis launched into the magnetic field forming an angle of 0º or 180º with the direction of the magnetic field lines.
In the figure above we see that the electric charge was launched at an angle θ = 0º and in the second situation it was launched at an angle θ = 180º. To determine the magnetic force acting on the particle we have the following equation:
F mg =|q|.vBsen θ
Since sin 0º = 0 and sin 180º = 0, we can conclude that no magnetic force will act on the load, that is, the force is zero. This gives us to understand that the particle is, inside the magnetic field, inuniform rectilinear motion .
Now let us suppose that the electric charge is launched perpendicularly, that is, forming an angle θ = 90º with the magnetic field lines.
According to the magnetic force equation,
F mg=|q|.vBsen θ
Since sin 90º = 1, we have that the magnetic force is:
F mg=|q|.vB
From the above equation we can conclude that the charge when launched perpendicularly to the lines of uniform magnetic field performs a uniform circular motion in a circle whose plane is perpendicular to the direction of the lines of the magnetic field.
