电荷在均匀磁场的运动

Suppose that we have a charge $q$ with mass $m$ and initial velocity $\vec{v}$ in a uniform magnetic field $\vec{B}$.

We would like to know as much as possible about the motion of the charge, including the trajectory of its motion and its speed at all times.

First, let us look at the speed.

It turns out that the speed of the charge never changes!

At first, this may seem very counterintuitive. However, let us recall how we have defined the Lorentz force: $$\vec{F}=q(\vec{v}\times\vec{B})$$ Note that the force is perpendicular to the velocity, meaning that the Lorentz force can never do any work. Therefore, the kinetic energy of the particle cannot change, and thus neither can its speed.

We have just mentioned that the Lorentz force is always perpendicular to the motion of the particle.

The centripetal force is always perpendicular to the velocity, just like our Lorentz force! Moreover, since we have already shown that the speed of the particle is constant, we can conclude that the radius of curvature of the particle will be constant.

Now, picture the shape of a curve that has a constant curvature.

It turns out that we could have three different possible curves: line, circle, and helix. Let us think about when these different cases would occur.

We will split the velocity of he particle into two components: $\vec{v}_{\parallel}$ and $\vec{v}_{\perp}$. These are the components of the velocity that are parallel and perpendicular to the magnetic field, respectively. Note that $\vec{v}_{\parallel}$ is unchanged by the Lorentz force.

If the particle is moving in a line, then from our earlier conclusion (speed is constant), the particle has no acceleration at all. This is only possible if the force is zero, meaning that $q(\vec{v}\times\vec{B})=0$. Thus, the particle moves in a line if and only if $\vec{v}\parallel\vec{B}$.

If the particle is moving in a circle, then no nonzero component of the particle's motion remains untouched. Thus, all of the particle's velocity is altered by the magnetic field. It follows that the particle moves in a circle if and only if $\vec{v}\perp\vec{B}$.

If the particle is moving in a circle, then a very important calculation is that of the radius of the circle. We can do this by equating the centripetal force and the Lorentz force: $$F_{mag}=F_{cent}$$ $$qvB=\frac{mv^2}{R}$$ $$R=\frac{mv}{qB}$$ This calculation of the radius is dimensionally correct, and it should agree with our intuition concerning how the different parameters affect the radius. Thus, we have confidence in our answer.

The last case, in which the velocity is neither parallel nor perpendicular to the magnetic field, results in one component of the velocity being unchanged (similar to the line), and one component constantly changed (similar to the circle). Thus, the particle moves in a helix in this last case. The radius of the helix is $\frac{mv_{\perp}}{qB}$, and the rate of ascent on the helix is $v_{\parallel}$.