# PHY5200 F07

## Chapter 2: Projectiles and Particles

Taylor 2.5-2.7 (today) and 3.1-3.2 (Friday)

### Motion of a Charge in a Uniform Magnetic Field

A charged particle moving through a magnetic field is subject to the Lorentz force

F = qv×B

where q is the particle's charge, v is the particle's velocity at a location r, and B is the magnetic field at location r. The magnetic field can change with position, but we will consider the relatively simple case when the magnetic field is constant, and choose the z axis of our coordinate system to be parallel to the magnetic field, B = Bzhat. This is an odd force, first because it is velocity dependent, and second because it is perpendicular to the direction of motion. But you are probably already familiar with the result for the motion of a charged particle in a uniform magnetic field. We'll look at how this problem is solved using techniques you've learned, with one additional twist.

The equation of motion for a charged particle moving in a uniform magnetic field is

mv[dot] = qv×B

The component equations are (assuming B points in the +z direction)

m dtvx = qBvy
m dtvy = -qBvx
m dtvz = 0

From the last equation we have vz = vz0 = constant. The x and y component equations are linear, coupled differential equaitons. The solution of these is an opportunity to demonstrate the method of change of variables.

Define ω = qB/m. The quantity ω is known as the cyclotron frequency, for reasons that are explained a little later. We can rewrite the x and y equations with ω as

dtvx = ωvy
dtvy = -ωvx

You should know from introductory physics that the motion in x-y is circular. To demonstrate that this is the case, and demonstrate the technique of change of variables to solve a differential equation, we write a new complex variable

η = vx + ivy.

The reason to choose this variable will become clear we we proceed -- experience led to this choice. I'll review complex numbers in a moment.

The time derivative of η is

dtη = dtvx + i dtvy = ωvy - iωvx = -iω(vx + ivy) = -iωη

With this choice of variable, we have gone from a pair of coupled differential equations to a single (complex) differential equation. Despite involving complex numbers, this equation is easily solved, as we've already done. The equation says that η is a function whose derivative is equal to the original function times a constant, and we've seen that this function is an exponential,

η = Ae-iωt.
straightforward except that the argument of the exponential is a complex number. We'll look at what it means to have a complex exponential.

### Complex Exponentials

The complex number i=(-1)½, so obviously i² = -1. An arbitrary complex number is written z = x + iy and can be represented in the complex plane, a two dimensional coordinate system where the real part of the number, x, is plotted along the horizontal (or x) axis and the imaginary part, y, is plotted along the vertical (or y) axis. Complex numbers can be added, subtracted, multiplied, and divided, as long as care is taken to keep track of the complex as well as the real part. Complex numbers resemble two-dimensional vectors in this sense.

Recall that the exponential function is equivalent to the infinite series

ez = 1 + z + z²/2! + z³/3! + ... = ∑n=0 zn/n!

and the derivative is

(d/dz)(Aekz) = k(Aekz).

These relations hold for complex numbers as well as real numbers.

But what does it mean to exponentiate a complex power? To investigate, it is useful to look at the simplest case of a complex exponential

eiθ = 1 + iθ + (iθ)²/2! + (iθ)³/3! + ...

This expression can be simplified using i² = -1, and grouping all even powers of i into a real part, and all odd powers into an imaginary part:

ei θ = [1 - θ²/2! + θ4/4! -...] + i [θ - θ³/3! + ...].

These two parts should look familiar: the real piece is the Taylor series for cosθ while the imaginary piece is i sinθ

eiθ = cosθ + isinθ.

This expression is known as Euler's formula. It is extremely useful in physics, and can be helpful to quickly (re)derive trigonometric relations such as for sine or cosine of a sum of angles.

Using the complex exponential, we can write the general solution as (even if A is a complex number)

η = Ae-iωt = aeiδe-iωt = aei(δ-ωt)

### Solution for the Charge in a B Field

With the solution for the velocity as a function of time, we can now solve for the position. To do that, it is convenient to define another complex variable, ξ, pronounced xi,

ξ = x + iy.

By taking the derivative of ξ we see that it equals η

dtξ = dtx + i dty = vx + i vy = η

so we can solve for ξ, and hence the positions x and y, by integrating η

ξ = ∫ ηdt = ∫ Ae-iωtdt = (iA/ω)e-iωt + constant.

Since x is the real part of ξ and y is the imaginary part, we can write, dropping the constant since it basically corresponds to a displacement of the motion:

x+iy = Ce-iωt

When t=0, the exponential is 1, so C corresponds to the position of the particle at that moment,

C = x0 + i y0

In general, the charge moves in a helix determined by the constant velocity in z and the circular motion in x-y. If the z velocity is zero, the motion reduces to a circle, centered on the origin. The radius of the circular motion is equal to the tangential velocity (x-y speed) divided by the angular velocity, ω (recall that v = rω)

r = v/ω = (mv)/(qB) = p/(qB).

This motion is the basis for many devices, in particular, the cyclotron.