Motion in a uniform magnetic field with finite E
When analysing the motion of a charged particle in a uniform magnetic field, it's essential to understand how different forces affect the particle's trajectory. Let's dive into the situation where there's also an electric field present.
Magnetic Field Only
For a particle with charge qqq and mass mmm moving in a uniform magnetic field B, the force on the particle is given by the Lorentz force law:
FB=q(v×B)
Here, v is the velocity of the particle. The magnetic force acts perpendicular to both the velocity of the particle and the magnetic field, leading to circular or helical motion depending on the initial conditions.
Adding an Electric Field
When there's also an electric field E present, the total Lorentz force on the particle becomes:
F=q(E+v×B)
The electric field contributes a force qEq \mathbf{E}qE that acts in the direction of E.
Equations of Motion
To find the particle's trajectory, we use Newton's second law:
mdv/dt=q(E+v×B)
Steady-State Solution
In a steady-state situation where the forces are balanced (assuming the particle is not accelerating), the velocity components reach constant values. For instance, if E and B are perpendicular:
If we allow electric field to be present, the motion will be the sum of two motions, the usual circular Larmor gyration plus a drift of the guiding center.
The equation of motion is now
Considering E to lie in the x z plane ( i.e Ey =0) and B= Bz, the z component of equation 1 is
dv_z/dt = q/m Ez
Integrating
The transverse components of equation 1 are
Equation 3 and 4 reduced to the previous case if we replace Vy by (Vy + Ex/B). Now solutions are
v_x = v⟂ e^iw_c t ...5
v_y = ± v⟂ e^iw_c t - Ex/ B ...6
Hence the Larmor motion is same before but there is superimpossed a drift V_gc of the guiding centre in the negative y axis for Ex> 0.
To obtain a general formula for Vgc, one cal solve equation 1 in vector form. We may omit the mdv/dt term in equation 1. Since this term gives only the circular motion at wc which we already know about. Then equation 1 becomes
E + vxB =0 ........7
Taking cross product with B
E x B = B x ( vxB) = vB² - B (v.B) ....8
The transverse components of this equation are
v_E = the electric field drift of guiding centre. It is independent of q,m and V⟂. The reason is obvious from the following physical picture. In the first half cycle of the ions orbit, it gains energy from the electric field and increases V⟂ snf hence r_L. In the second half cycle, it loses energy and decreases in r_L. This difference in r_L on the left and right sides of the orbit causes the drift V_E. For gravitational field, since F = qE, the guiding centre drift caused by F is then
V_f = 1/ q F xB/ B² ...10
In particular if F is the force of gravity mg there is a drift
V_g = m/q gx B/ B² ....11
This is similar to V_E in the sense that it is also perpendicular to both the force and B but it is different in one important aspect. The gravitational drift causes electrons and ions to drift in opposite directions due to their opposite charges. This results in a net current density in the plasma, as the contributions from ions and electrons add up in the direction of the gravitational force.
Summary
In a uniform magnetic field with an electric field present, the motion of a charged particle is affected by both fields. The electric field modifies the particle's trajectory compared to the case with only a magnetic field. Specifically:
The particle’s motion can be described as a combination of circular or helical motion due to the magnetic field and a uniform drift in the direction of E×B.
The z-component of motion will be affected directly by Ez if present.
Understanding these interactions allows you to predict and analyze the behaviour of charged particles in various electromagnetic environments.
This note is a part of the Physics Repository