# Magnetic field (correction term)

## Homework Statement:

I have attached my attempt, I have a lof doubts to go to the next step to solve the problem

## Relevant Equations:

Ampere´'s law

Ampere´'s law with the correction term
I have a infinite cylinder with radius R with a current density ,

and magnetic field

.

I have to proof that it is acceptable to discard the correction term of term of ampere's law, while calculating the magnetic field, as long as it obeys the following condition:

I cant seem to manipulate the equation of the magnetic field to actually come to the conclusion above. I know how to calculate the magnetic field, and I know I didnt consider the correction term dE/dt, I just dont understand well how to proof it like it is asked.

Attachments

#### Attachments

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etotheipi and Delta2

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Delta2
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Two questions
1) Is the current density only in the surface of the cylinder or it is also in the interior of the cylinder? If I judge by the formula you give for B in the third picture it must be a surface current density (only in the surface of the cylinder).
2) Are Jeffimenko's equations within the scope of your course material?
https://en.wikipedia.org/wiki/Jefimenko's_equations

Also the pdf file you attach the writings are fainted out a lot and it is very hard to read.

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etotheipi
Delta2
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Gold Member

I am not 100% sure if the expressions for the calculated Electric field are correct. They look OK.

However I believe for the second part it is better to work with Ampere's Law in integral form. Take as amperian loop a rectangular loop that has one side along the z-axis and at the center of the cylinder and the other side along the y (or x) axis (or in general along the radial direction) and the length of this side equal to R(radius of the cylinder).
I think if you apply Ampere's law in integral form as I suggest you ll end up with an equation similar to $$B\cdot l=\mu_0k(t)l+\mu_0\epsilon_0\int_S\frac{dE}{dt}dS$$ where $l$ is the length of the amperian loop along the z-axis and $S=R\cdot l$ is the surface area of the amperian loop. Your job will be to evaluate that integral and then do some basic algebra to reach the desired conclusion.

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I am not 100% sure if the expressions for the calculated Electric field are correct. They look OK.

However I believe for the second part it is better to work with Ampere's Law in integral form. Take as amperian loop a rectangular loop that has one side along the z-axis and at the center of the cylinder and the other side along the y (or x) axis (or in general along the radial direction) and the length of this side equal to R(radius of the cylinder).
I think if you apply Ampere's law in integral form as I suggest you ll end up with an equation similar to $$B\cdot l=\mu_0k(t)l+\mu_0\epsilon_0\int_S\frac{dE}{dt}dS$$ where $l$ is the length of the amperian loop along the z-axis and $S=R\cdot l$ is the surface area of the amperian loop. Your job will be to evaluate that integral and then do some basic algebra to reach the desired conclusion.
Hi, thanks for answering I have a few questions about what you said, I have attached a new file with a few additions for you see what I am doing, I cant seem to understand a few things.

#### Attachments

• 457.7 KB Views: 5
Two questions
1) Is the current density only in the surface of the cylinder or it is also in the interior of the cylinder? If I judge by the formula you give for B in the third picture it must be a surface current density (only in the surface of the cylinder).
2) Are Jeffimenko's equations within the scope of your course material?
https://en.wikipedia.org/wiki/Jefimenko's_equations

Also the pdf file you attach the writings are fainted out a lot and it is very hard to read.
I have attached a new file with more information, I hope it clarifies your first question, I havent study jeffimenko equations, it wasnt mentioned in much detail in my electromagnetism course.

#### Attachments

• 457.7 KB Views: 5
Delta2
Homework Helper
Gold Member
Its ok Jeffimenko's equations involve retarded times and are good usually when the current /charge densities are bounded by finite volume.

Answering your first question it is actually $$dS=dzd\rho$$ where$\rho$ the radial cylindrical coordinate as it appears in your expression for the electric field $$\bf{E}=-\frac{\mu_0\dot k(t)\rho}{2}\hat \phi$$, for $\rho\leq R$. (You are using a weird symbol instead $\rho$ is that symbol supposed to be just $r$?)

The area S on which we are integrating is rectangular as i said. We have a double integral with boundaries of integration from 0 to L (for the integration with respect to dz) and from 0 to R (for the integration with respect to $d\rho$.

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yes ρ is r, is basically the radius, it is the notation I am used from my course, but it is the same thing. So after I calculate the integral , what manipulation I need to do to come to conclusion that condition 2 in my file is valid, even when I isolate the ratio of the second derivate of density of current and density of current, I dont know what conclusion to draw from there.
Also you said ds= dL dρ which would make sense if it was the magnetic flux but in the equation the term of correction is derivate of time of the electric flux, which means dS shouldnt be just the loop where the current is flowing. I am sorry I am not trying to correct you or anything like that, it is just something that really confuses me, sometimes I get stuck in simple details.

Delta2
Homework Helper
Gold Member
if you calculate the integral correctly and do the algebra correctly you ll get
$$B=\mu_0k(t)(1-\frac{R^2\ddot k(t)}{4c^2k(t)})$$
Now what do you think the condition should be so that you can ignore the term inside the parenthesis

Delta2
Homework Helper
Gold Member
Maybe if I say $$dS=dzd\rho$$ will it be better for your understanding? I said that the amperian loop is a rectangular such that one of its sides (the one with length L) is along the z-axis and its other side is along the radial axis and has length R(=radius of cylinder). It is obvious to me that the electric field having the $\hat\phi$ direction is always perpendicular to the surface of the amperian loop.

I don't know where exactly your misunderstanding lies , but the amperian loop I chose , is not a circular loop around the outer surface of the cylinder.

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