How Does Angular Frequency Impact Wave Calculations in Electromagnetic Theory?

[RememberYourAngles]

Homework Statement

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Ignore the suggested Problems if you will. If you can't see the image give me a shout and I'll give the problem statement here.

Homework Equations

B(y,t)=Bmaxcos(Ky-Wt)
Wavelength=2pi/kn
W=2pi*Frequency
V=Emax/Bmax
V=Walength*Frequency

The Attempt at a Solution

So calculating the Wavelength is no problem. We get .00024166m for the Wavelength
The problem comes when we try to get the frequency from the stated Angular Frequency, which strangely seems to be multiplied by the Bmax to attain a unit of Tesla Inverse Seconds. Is it? And if it is does that affect our calculation for Wavelength?

 

[NFuller]

So calculating the Wavelength is no problem. We get .00024166m for the Wavelength
The problem comes when we try to get the frequency from the stated Angular Frequency, which strangely seems to be multiplied by the Bmax to attain a unit of Tesla Inverse Seconds. Is it? And if it is does that affect our calculation for Wavelength?

The question doesn't ask for the wavelength. The frequency is the same for both the electric and magnetic component of the wave. The relationship between the electric and magnetic component of a plane wave is relatively simple to derive using the Maxwell-Faraday equation
$$\nabla\times\mathbf{E}=-\frac{\partial\mathbf{B}}{\partial t}$$
Since the electric field is parallel to both the magnetic field (in the x-direction here) and the direction of wave propagation (in the y-direction here), the above equation reduces to
$$\frac{\partial E_{z}}{\partial y}=-\frac{\partial B_{x}}{\partial t}$$
I assume a similar relation is in your textbook somewhere. Can you use this to find $\mathbf{E}$?

 

[RememberYourAngles]

I imagine you can...maybe...but I don't know how to, nor does the textbook go over how to derive a wave function from something like that.
I was already aware of the directional components so that's not terrible helpful.
We're asked to find the wave function of the Electric Field component or E(y,t) Not just E. Regardless to find E we'd need to know the velocity of the wave and we can't assume its just c because we don't know the medium or source of the Em wave.
The idea was to use the wavelength and frequency to find the velocity of the wave, then use the velocity to find the Amplitude of the Electric Field, and then we'd have everything we'd need to construct the wave function.

If I wanted to attempt the problem this way It'd mean that the B field as it varies with time is equivalent to the negative of the E field as it varies with distance?

 

[NFuller]

Regardless to find E we'd need to velocity of the wave

Not really. The frequency and wavelength are the same for both the electric and magnetic field. The part that is different is the amplitude.

I imagine you can...maybe...but I don't know how to, nor does the textbook go over how to derive a wave function from something like that.

Really, I'm a bit surprised the text would not have this. Have you taken calculus yet?

 

[RememberYourAngles]

Not really. The frequency and wavelength are the same for both the electric and magnetic field. The part that is different is the amplitude.

Really, I'm a bit surprised the text would not have this. Have you taken calculus yet?

The question doesn't ask for the wavelength. The frequency is the same for both the electric and magnetic component of the wave. The relationship between the electric and magnetic component of a plane wave is relatively simple to derive using the Maxwell-Faraday equation
$$\nabla\times\mathbf{E}=-\frac{\partial\mathbf{B}}{\partial t}$$
Since the electric field is parallel to both the magnetic field (in the x-direction here) and the direction of wave propagation (in the y-direction here), the above equation reduces to
$$\frac{\partial E_{z}}{\partial y}=-\frac{\partial B_{x}}{\partial t}$$
I assume a similar relation is in your textbook somewhere. Can you use this to find $\mathbf{E}$?

$$\frac{\partial E_{z}}{\partial y}=-\frac{\partial B_{x}}{\partial t}$$
Alright so this means that the time dependent component of the Magnetic Field is equivalent to the negative of the distance dependent component of the Electric Field?

Regardless can you answer me this: The time dependent component of the electric field we're given is the angular frequency multiplied by the amplitude of the B field?

 

[NFuller]

Alright so this means that the time dependent component of the Magnetic Field is equivalent to the distance dependent component of the Electric Field?

Their derivatives are equivalent. To start, can you work out the derivative on the right hand side of the equation using your formula for B?

 

[RememberYourAngles]

Alright so If we take the partial derivative with respect to time of the wave function of the B field, we're left only with the distance dependent component.
And this is equivalent to the negative of the partial derivative with respect to distance of the wave function of the E field, which leaves us only with the time dependent component.
Is this an accurate statement?

 

[RememberYourAngles]

So the right hand side gives us -2.32323sin((26000m^-1)y+((.897Ts^-1)t)I hat
That gives us the partial derivative with respect to distance of the E field.
What do we do from here?
Is there such a thing as a partial integral?

 

[NFuller]

Alright so If we take the partial derivative with respect to time of the wave function of the B field, we're left only with the distance dependent component.

No, time will still be in the argument of the trig function.

And this is equivalent to the partial derivative with respect to distance of the of the wave function of the E field, which leaves us only with the time dependent component.

No, distance will be here too. It appears that you are not very familiar with partial differential equations so I will do the right hand side.
$$-\frac{\partial B_{x}}{\partial t}=(2.59\mu T)(0.897Ts^{-1})\text{sin}((26km^{-1})y+(0.897Ts^{-1})t)=(2.32MT/s)\text{sin}((26km^{-1})y+(0.897Ts^{-1})t)$$

 

[RememberYourAngles]

No, time will still be in the argument of the trig function.

No, distance will be here too. It appears that you are not very familiar with partial differential equations so I will do the right hand side.
$$-\frac{\partial B_{x}}{\partial t}=(2.59\mu T)(0.897Ts^{-1})\text{sin}((26km^{-1})y+(0.897Ts^{-1})t)=(2.32MT/s)\text{sin}((26km^{-1})y+(0.897Ts^{-1})t)$$

Alright so do we take a partial integral with respect to distance to get the E equation?

 

[NFuller]

So the right hand side gives us -2.32323sin((26000m^-1)y+((.897Ts^-1)t)I hat

Ok great, aside from missing units that matches my answer.

Is there such a thing as a partial integral?

You can just integrate this with respect to y now.

 

[RememberYourAngles]

Ok great, aside from missing units that matches my answer.

You can just integrate this with respect to y now.

I appreciate the help bro.
I have to admit I'm a tiny bit aggravated at teach because none of the examples using the wave function dealt with actually using any of the relationships in terms of partial derivatives

 

[NFuller]

I have to admit I'm a tiny bit aggravated at teach because none of the examples using the wave function dealt with actually using any of the relationships in terms of partial derivatives

I am a bit curious how he expects this to make sense without using calculus. Maybe there is a canned equation he gave you somewhere. Also, let us know your final answer when you get it so it can help others on the forum.

 

[RememberYourAngles]

I am a bit curious how he expects this to make sense without using calculus. Maybe there is a canned equation he gave you somewhere. Also, let us know your final answer when you get it so it can help others on the forum.

Naw he did give us that relationship.
Its just that none of the examples either in class or from the suggested problems used it, meaning most people are liable to forget it exists when it comes time to solving problems.
The fact that its lack of use in the suggested problems also means it won't be playing a part in the test also helps to let people ignore it.
I do have one last question. Shouldn't the prefix be 2.32 micro tesla? Not 2.32 mega tesla?

 

[NFuller]

Shouldn't the prefix be 2.32 micro tesla? Not 2.32 mega tesla?

No because we are multiplying the prefix $\mu=10^{-6}$ with the prefix $T=10^{12}$ which equals $M=10^{6}$.

 

[RememberYourAngles]

No because we are multiplying the prefix $\mu=10^{-6}$ with the prefix $T=10^{12}$ which equals $M=10^{6}$.

Wait T was a prefix and not Tesla?
Oh right T=Tera
Man that means my original idea would've worked.
Still. Thanks.

 

[kuruman]

I am a bit curious how he expects this to make sense without using calculus. Maybe there is a canned equation he gave you somewhere. Also, let us know your final answer when you get it so it can help others on the forum.

The given magnetic field is of the form $$\vec{B}(y,t) = B_0 ~\hat{i}\cos((k~y+\omega~t)$$It represents an EM wave traveling in the negative y-direction as seen by the relative "+" sign in the phase. The associated electric field is in phase with the magnetic field, it has magnitude $c B_0$, it is orthogonal to the B-field such that $\vec{E} \times \vec{B}$ is in the direction of propagation, in this case $\hat{j}$. Putting it all together gives $$\vec{E}(y,t) = -cB_0 ~\hat{k}\cos((k~y+\omega~t)$$
All this requires elementary understanding of EM wave properties without need for calculus.