Find the phase velocity of the wave (Plasma Physics)

Firben
Messages
141
Reaction score
0

Homework Statement


An electromagnetic wave with the frequency f = 1.4 GHz is propagating in the z-direction from vacuum in a plasma with the magnetic field B_0 = 0.1T z^. The plasma density is 1.0*10^17 particles/m^3. The wave is the separated into a R- and L-wave in the plasma.

Homework Equations


w_c = |q|*B/m cyclotron angular resonance
w_p = (n_0*e^2/(ε*m_e))^(1/2) plasma angular frequency
(ck/w)^2 = 1-(w_p^2*w^2)/(1-(w_c/w)) R-Wave (whistler)
(ck/w)^2 = 1-(w_p^2*w^2)/(1+(w_c/w)) L-wave

(v_ph)_R = (w/ck)_R = ((w(w-w_c))/(w^2-w*w_c-w_p^2))^(1/2)
(v_ph)_L = (w/ck)_R = ((w(w+w_c))/(w^2+w*w_c+w_p^2))^(1/2)

B_0 = 0.1T z^
f = 1.4 GHz = 8.8*10^9 rad/s
w_p = 1.78*10^10 rad/s
w_c = 1.76 * 10^10 rad/s

The Attempt at a Solution



Inserting these values into the formula for the phase velocity, i get the following

(v_ph)_R = c*0.443 = 1.329 * 10^8 m/s,
since the phase velocity is defined as w/k = v_ph. So i multiple it with c
correct value is v_ph = 6.6*10^6 m/s
(v_ph)_L = no wave, since it is imaginary

Then i want to know which polarization the wave has after z=1 m
 
Physics news on Phys.org
Can you write it with in LaTeX?
 

Homework Statement


An electromagnetic wave with the frequency f = 1.4 GHz is propagating in the z-direction from vacuum in a plasma with the magnetic field ## B_0 = 0.1T \hat z ##. The plasma density is ## 1.0*10^{17} ## ## \frac {particles} {m^3} ##. The wave is then separated into a R- and L-wave in the plasma.

2. Homework Equations
## w_c = \frac {|q|*B} m ## cyclotron angular resonance

## w_p = \sqrt \frac {n_0*e^2 } {ε*m_e} ## plasma angular frequency

## (\frac {ck} {w})^2 = 1- \frac {w_p^2/w^2} {1-(w_c/w)} ## R-Wave (whistler)
## (\frac {ck} {w})^2 = 1- \frac {w_p^2/w^2 } {1+(w_c/w)} ## L-wave

Phase Velocity:

## (v_{ph})_R = (\frac {w} {ck})_R = \sqrt \frac {w(w-w_c) } {w^2-w*w_c-w_p^2} ##
## (v_{ph})_L = (\frac {w} {ck})_L = \sqrt \frac {w(w+w_c) } {w^2+w*w_c-w_p^2 } ##

## B_0 = 0.1T \hat z ##
## f = 1.4 GHz = 8.8*10^9 ## rad/s
##w_p = 1.78*10^{10} ##rad/s
##w_c = 1.76 * 10^{10} ## rad/s

3. The Attempt at a Solution

Inserting these values into the formula for the phase velocity, then i got the following

## (v_{ph})_R = c*0.443 = 1.329 * 10^8 ##m/s,
since the phase velocity is defined as ## \frac {w} {k} = v_{ph} ##. So i multiple it with c.
The correct value is ## v_{ph} = 6.6*10^6 ##m/s
##(v_{ph})_L ## = no wave, since it is imaginary

How can i know which polarization the wave has after z=1 m ?
 
Why is my solution wrong ?
 
Im still stuck, why do i get a different answer ?
 
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
Back
Top