How Does the Magnetic Field Relate to Frequency in Electromagnetic Waves?

[hidemi]

Homework Statement

A sinusoidal emf is connected to a parallel plate capacitor. The magnetic field between the plates is:

A. 0
B. constant
C. sinusoidal and its amplitude does not depend on the frequency of the source
D. sinusoidal and its amplitude is proportional to the frequency of the source
E. sinusoidal and its amplitude is inversely proportional to the frequency of the source

The answer is D

Relevant Equations

(See better expression below)

https://www.physicsforums.com/attachments/282201
Are we using this equation above to explain this question? The magnetic field is definitely in sinusoidal form but how does it proportional to the frequency of the source?

 

[haruspex]

Your attachment is unreachable for me.
As the charge varies, what is happening to the E field? What does that mean for the B field?

 

[hidemi]

Your attachment is unreachable for me.
As the charge varies, what is happening to the E field? What does that mean for the B field?

I attached again.
The E field also varies and so does the B field. Are you trying to refer their relationship to the wave function of Maxwell equations?
E = Em*sin(kx-wt)
B = Bm*sin(kx-wt)

 

[Delta2]

There will be electromagnetic waves from the capacitor whose voltage varies sinusoidal but they won't be plane waves like those equations you post at #3.
To find the form of waves assume that the electric field between the plates of the capacitor varies only with time but not with space, so it has the formula $$E=\frac{V}{l}=\frac{V_0}{l}\sin\omega t$$ where $l$ the distance between the capacitor's plates. Then use Ampere-Maxwell law to find the magnetic field generated by such an electric field. You can assume that the magnetic field lines will be concentric circles with their plane parallel to the plane of the capacitor's plates.

 

[hidemi]

There will be electromagnetic waves from the capacitor whose voltage varies sinusoidal but they won't be plane waves like those equations you post at #3.
To find the form of waves assume that the electric field between the plates of the capacitor varies only with time but not with space, so it has the formula $$E=\frac{V}{l}=\frac{V_0}{l}\sin\omega t$$ where $l$ the distance between the capacitor's plates. Then use Ampere-Maxwell law to find the magnetic field generated by such an electric field. You can assume that the magnetic field lines will be concentric circles with their plane parallel to the plane of the capacitor's plates.

Thank you for the explanation!

 

[Delta2]

Thank you for the explanation!

Were you able to derive why D. is true using my suggestion at post #4?

 

[hidemi]

Were you able to derive why D. is true using my suggestion at post #4?

E = ω/k * B
As E varies, B does so based on their correlation. According to the formula in post #4, it illustrates how E relates to ω which is generated by the source. Is this you are looking for?

 

[haruspex]

E = ω/k * B

I don't think that's quite what Ampere's law says. Should be a derivative in there.
But in terms of amplitudes over the cycle, yes.

 

[Delta2]

No this isn't exactly what i had in mind. What I had in mind is to apply Ampere-Maxwell law to calculate B for the E as given in post #4.

 

[hidemi]

No this isn't exactly what i had in mind. What I had in mind is to apply Ampere-Maxwell law to calculate B for the E as given in post #4.

I think:
B * 2πr = μ0*ε0*dE/dt*πr^2
B = μ0*ε0*[v/l * cos(wt)] /2

 

[Steve4Physics]

B * 2πr = μ0*ε0*dE/dt*πr^2
B = μ0*ε0*[v/l * cos(wt)] /2

Maybe @haruspex and @Delta2 have gone to bed! So I'll push in!

Your first equation (above) looks correct. You're getting really close! But you don't need to assume the plates are circular - you could replace $\pi r^2$ with A (area).

(I'll ignore your second equation as it's incorrect - you haven't differentiated properly. Also you have introduced (I think) the max. voltage and the plate separation, which are unnecessary; stick with $E_0$.)

Using $E = E_0 \sin( \omega t)$, differentiate (correctly!) to find dE/dt. Put this into your first equation. Now you can get an equation for B. Look at this and see if you can answer the original question based on this equation.

If using angular frequency ($\omega$) is causing any confusion, remember $f = \frac {\omega}{2\pi}$. Using (temporal) frequency (f) or angular frequency ($\omega$) doesn't change the correct choice of answer.

 

[hidemi]

Using $E = E_0 \sin( \omega t)$, differentiate (correctly!) to find dE/dt. Put this into your first equation. Now you can get an equation for B. Look at this and see if you can answer the original question based on this equation.

B*2πr = μ0 * ε0* (E0*ω*cos(ωt))
So, this equation tells the magnetic field is sinusoidal and its amplitude (E0*ω) is proportional to the frequency of the source.

 

[Steve4Physics]

The amplitude of the B-field is not "E_0*ω".

If you want the amplitude ($B_0$) rearrange your formula into the form:
B = (expression)cos(ωt)

Then (expression) is $B_0$.

However, you are correct in saying "amplitude ... is proportional to the frequency of the source". So this answers the original (Post #1) question!

 

[hidemi]

The amplitude of the B-field is not "E_0*ω".

If you want the amplitude ($B_0$) rearrange your formula into the form:
B = (expression)cos(ωt)

Then (expression) is $B_0$.

However, you are correct in saying "amplitude ... is proportional to the frequency of the source". So this answers the original (Post #1) question!

Thanks for your further clarification and explanation.

 

[rude man]

Instead of working with fields I would work with the displacement current which is of course $\omegaCV$; assume uniform current density along the plates, then use Ampere's law for the B field inside (& outside for extra credit) the capacitor.

BTW this is beyond intro physics but in fact as you increase the frequency more & more you get more and more new E and B fields.

 

[Delta2]

BTW this is beyond intro physics but in fact as you increase the frequency more & more you get more and more new E and B fields.

yes i agree the full answer to that lies in Feynman lecture and is all about cavity resonator concept. Perhaps you can post the link to the Feynman Lectures if the OP is interested to read more about this