Thornton and Marion, Chapter 2, Prob. 22

[union68]

Homework Statement

This is a four part problem -- the only issue I have is on part (c) so I'll condense the question:

If the equation of motion of a particle of charge $$q$$ in an electric field and magnetic field is

$$qv_y B \hat{\mathbf{i}} + \left( qE_y - qv_xB \right) \hat{\mathbf{j}} + qE_z \hat{\mathbf{k}} = m \frac{d}{dt} \left( v_x \hat{\mathbf{i}} +v_y \hat{\mathbf{j}} + v_z \hat{\mathbf{k}}\right),$$

obtain expressions for $$v_x \left(t\right)$$ and $$v_y \left(t\right)$$. Show that the time averages of these velocity components are

$$\left\langle v_x\left(t\right)\right\rangle = \frac{E_y}{B}$$

and

$$\left\langle v_y\left(t\right)\right\rangle = 0$$

(Show that the motion is periodic and then average over one complete period.)

Homework Equations

The Attempt at a Solution

Solutions to the DE are

$$v_x \left(t\right) = \frac{E_y}{B} + C_1 \cos \left(\omega t\right) + C_2 \sin \left(\omega t\right)$$

and

$$v_y \left(t\right) = -C_1 \sin \left(\omega t\right) + C_2 \cos \left(\omega t\right)$$,

where $$\omega$$ is the cyclotron frequency. Now, clearly the velocity functions are periodic with period $$2\pi/\omega$$. What do they mean "time averages," and what does the $$\left\langle \right\rangle$$ notation mean?

Perhaps,

$$\left\langle v_x\left(t\right)\right\rangle = \frac{ v_x\left(0\right) + v_x \left(2\pi/\omega\right)}{2}$$,

like an arithmetic mean, but I'm just guessing based on the hint (and that doesn't yield the correct answer). Do they mean average velocity?

 

[tiny-tim]

Hi union68!

Solutions to the DE are

$$v_x \left(t\right) = \frac{E_y}{B} + C_1 \cos \left(\omega t\right) + C_2 \sin \left(\omega t\right)$$

and

$$v_y \left(t\right) = -C_1 \sin \left(\omega t\right) + C_2 \cos \left(\omega t\right)$$,

where $$\omega$$ is the cyclotron frequency. Now, clearly the velocity functions are periodic with period $$2\pi/\omega$$. What do they mean "time averages," and what does the $$\left\langle \right\rangle$$ notation mean?

"Time average" means the average value over time (and the <> notation just means average, or expectation value) …

in this case, theoretically you'd integrate over a whole period, and then divide by the period.

However, in this case it's a bit obvious that the average of a cos or sin over the period is 0, so that just leaves you with … ?

 

[union68]

Ah, yes. Integrating over the period and dividing by the period gives you the correct answer. The cosines and sines vanish.

However, what exactly is this procedure though? I guess I've never seen integration related to an average.

 

[tiny-tim]

However, what exactly is this procedure though? I guess I've never seen integration related to an average.

uhh? … how else would you define an average of something that varies continuously?

(and how do you think root-mean-square voltage etc is defined? )

 

[union68]

After a quick Google search I found some explanation. I have never seen continuous averaging before.

I have no idea what RMS voltage is either -- I'm a math major dabbling in physics.

Thank you for the help.