Magnetic field in a rotating uniformly-charged infinite cylinder

[sagigever]

Homework Statement

inside a cylinder (not solid) with radius $$a$$ and infinite lengh there is a charge with uniform density $$\rho$$. the charge rotating inside the cylinder in speed $$\omega(r)=\omega_0\frac{a}{r}$$
find the vector of magnetic field inside the cylinder

Relevant Equations

$$\sum B_{\parallel} \cdot \Delta l = \mu_0 \ I_{enclosed} \ \text{, } \ \ \ \text{ or } \ \ \ \oint \vec{B} \cdot \vec{dr} = \mu_0 \ I_{enclosed}$$

I am sure I need to use Amper's law to do that. if I use the equation I mentioned above it easy to calculate the right side of the equation but I have problem how to calculate the path integral.
I know from right hand rule that the magnetic field will point at $$Z$$ and the current is in $$\phi$$ direction, so I want to take a loop of rectangle
but than I have no Idea how to calculate because $$\vec{B}_(x,y,z) not equal to \vec{B}_(0,0,z)$$

I hope I was clear enough to describe excatly where I am stuck

 

[Delta2]

What symmetries do you think are present in this problem?

 

[sagigever]

What symmetries do you think are present in this problem?

Cylinder, in addition I also know the magnetic field of hollow cylinder but when I try to use this technique here it is not working, ofc because the current inside the cylinder is not zero

 

[Delta2]

Can you be a little more precise with describing the symmetry? In a cylindrical coordinate system, how do you think the magnetic field varies with respect to $z$, $\rho$, $\phi$?

 

[sagigever]

Can you be a little more precise with describing the symmetry? In a cylindrical coordinate system, how do you think the magnetic field varies with respect to $z$, $\rho$, $\phi$?

ohh ok, I can see that according to right hand rule the magnetic field will be pointing at $$\hat z$$ and also my current is in the direction $$\hat \phi$$. so I guess I can take out the magnetic field out from the integral? because the cylinder is infintie

 

[Delta2]

You are right that the magnetic field will be in the z-direction. Also right that the current is in $\hat\phi$.
But you are not answering my question about the symmetries present in this problem
Does the magnetic field varies along the z-axis?
Does the magnetic field varies along the radial-axis?
Does the magnetic field varies as the $\phi$ varies?

 

[sagigever]

You are right that the magnetic field will be in the z-direction. Also right that the current is in $\hat\phi$.
But you are not answering my question about the symmetries present in this problem
Does the magnetic field varies along the z-axis?
Does the magnetic field varies along the radial-axis?
Does the magnetic field varies as the $\phi$ varies?

I think in the z-axis it will stay constant and on the radial it will vary, this is because we have a charge with angular speed depending on the distance from the center. I think there is no componet in the $\phi$ axis

 

[Delta2]

I think in the z-axis it will stay constant and on the radial it will vary, this is because we have a charge with angular speed depending on the distance from the center. I think there is no componet in the $\phi$ axis

Correct regarding the 2/3 of the question. (Also correct that there is no component in the $\hat\phi$ direction). But do you think that the magnetic field has azimuthal ($\phi$) symmetry? Or does it varies as $\phi$ varies?

 

[sagigever]

Correct regarding the 2/3 of the question. (Also correct that there is no component in the $\hat\phi$ direction). But do you think that the magnetic field has azimuthal ($\phi$) symmetry? Or does it varies as $\phi$ varies?

I am not sure how to aprroach this question. if $\phi$ so I think the magnetic field will vary but this is only intuition

 

[Delta2]

We disagree here, I think the B-field has azimuthal symmetry but let's continue and see how this issue will automatically be addressed when we calculate the line integral of B vector and the surface integral of the current density vector.
For what follows i use the cylindrical coordinates $(\rho,\phi,z)$.
Take a rectangular loop in which you going to apply Ampere's law as follows:
From point $(0,0,0)$ to point $(0,0,L)$ then to point $(r,0,L)$ down to point $(r,0,0)$ and back to point $(0,0,0)$. (I should ve made a drawing here but I am really bad using drawing programs)
How do you evaluate $\int \vec{B}\cdot d\vec{r}$ along that loop?
How are you going to evaluate the total current that is enclosed by this loop?

 

[sagigever]

We disagree here, I think the B-field has azimuthal symmetry but let's continue and see how this issue will automatically be addressed when we calculate the line integral of B vector and the surface integral of the current density vector.
For what follows i use the cylindrical coordinates $(\rho,\phi,z)$.
Take a rectangular loop in which you going to apply Ampere's law as follows:
From point $(0,0,0)$ to point $(0,0,L)$ then to point $(r,0,L)$ down to point $(r,0,0)$ and back to point $(0,0,0)$. (I should ve made a drawing here but I am really bad using drawing programs)
How do you evaluate $\int \vec{B}\cdot d\vec{r}$ along that loop?
How are you going to evaluate the total current that is enclosed by this loop?

this is exactly what I did from the beginning and maybe I wasn't clear enough but I don't know how to calculate

$\int \vec{B}\cdot d\vec{r}$ , I know that both path from $(0,0,L)$ to $(r,0,L)$ and vise verse will cancel each other but nothing more than that

 

[Delta2]

It is not wrong to say that they will cancel each other but it is more precise to say that it would be zero in these paths (why? hint the B-field is always in the z direction).
Along the straight path from $(0,0,0)$ to $(0,0,L)$ isn't it $B(0,0)L$? (why?)
Along the straight path from $(r,0,L)$ down to $(r,0,0)$ isn't it $B(r,0)L$?
What do you think?

 

[sagigever]

It is not wrong to say that they will cancel each other but it is more precise to say that it would be zero in these paths (why? hint the B-field is always in the z direction).
Along the straight path from $(0,0,0)$ to $(0,0,L)$ isn't it $B(0,0)L$? (why?)
Along the straight path from $(r,0,L)$ down to $(r,0,0)$ isn't it $B(r,0)L$?
What do you think?

yes for both qeustions, because the field is constant in the z direction?

 

[sagigever]

It is not wrong to say that they will cancel each other but it is more precise to say that it would be zero in these paths (why? hint the B-field is always in the z direction).
Along the straight path from $(0,0,0)$ to $(0,0,L)$ isn't it $B(0,0)L$? (why?)
Along the straight path from $(r,0,L)$ down to $(r,0,0)$ isn't it $B(r,0)L$?
What do you think?

so how do I know that value of $B(r,0)$?

 

[Delta2]

you don't know it, you ll find it if you solve the equation of Ampere's law. You said that you can calculate easily the right hand side of Ampere's law (that is the current density surface integral of the rectangular loop).

 

[Delta2]

So the left hand side of Ampere's law ( the path integral) is equal to $(B(0,0)+B(r,0))L$. What is the right hand side equal to?

 

[sagigever]

So the left hand side of Ampere's law ( the path integral) is equal to $(B(0,0)+B(r,0))L$. What is the right hand side equal to?

Yes but I have

$(B(0,0)+B(r,0))L$

I need only B

 

[Delta2]

You can consider $B(0,0)$ to be like a constant (this constant turns out to be zero). Solve the equation of Ampere's law for $B(r,0)$.

 

[sagigever]

You can consider $B(0,0)$ to be like a constant (this constant turns out to be zero). Solve the equation of Ampere's law for $B(r,0)$.

okk that's the thing I was looking for, can you explain why $B(0,0)$ is zero?

 

[Delta2]

yes well I got myself confused as well here, I think it is zero because of the azimuthal symmetry of the current density of this system. It lies right at the center of rotation and all the contributions from the surrounding rotating currents are getting canceled there.

 

[sagigever]

yes well I got myself confused as well here, I think it is zero because of the azimuthal symmetry of the current density of this system. It lies right at the center of rotation and all the contributions from the surrounding rotating currents are getting canceled there.

I am little bit confused now, is there only one comopnent of the magnetic field in the $Z$ axis?

 

[Delta2]

There are no more components (along $\hat\phi$ or along $\hat\rho$), only along $\hat z$. Why you saying that? Can you post what is the final expression you get for $B(r,0)$?

 

[sagigever]

There are no more components (along $\hat\phi$ or along $\hat\rho$), only along $\hat z$. Why you saying that? Can you post what is the final expression you get for $B(r,0)$?

Yes, I got $\rho \omega_0 ar \hat z$

 

[sagigever]

I am asking that because the second question is asking about relativity of magnetic field, it says that a person who moves speed $V$ perpendicular to the axis of cylinder, in his perspective he messures at point d(inside the cylinder) magntic field zero. I am asked to find $d$. so I need to use field transformations but if I have no component of magnetic field in other direction it mean d=0 and I don't think it is possible

 

[Delta2]

That's what I got also. You can take the loop to be for any $\phi$ (from $(0,\phi,0) \to (0,\phi,L) \to(r,\phi,L) \to (r,\phi,0)$ and nothing changes you get the same result so indeed the B-field has azimuthal symmetry (it is independent of $\phi$).

Sorry i will not be able to comment on the relativity question cause i am pretty bad with relativity (special and general).

 

[Delta2]

Only mistake I can see is that $B(0,0)$ is not necessarily zero, maybe B is at maximum there.

 

[sagigever]

Only mistake I can see is that $B(0,0)$ is not necessarily zero, maybe B is at maximum there.

I know for sure that the magnetic field on the symmetry axis of hollow cylinder with surface charge density is $u_0 k \hat z$ is that true if I will say the same for this cylinder and just change the $k$ to $J$ and than I can express $J$ as $\omega r \rho$

 

[Delta2]

I think we might have a slight blunder here, probably the correct formula is $$B(r)=(B(0)-\mu_0\rho\omega_0\alpha r)\hat z$$

 

[sagigever]

I think we might have a slight blunder here, probably the correct formula is $$B(r)=(B(0)-\rho\omega_0\alpha r)\hat z$$

Yes I think too, what do you think about my idea to express $B(0)$ with the result for $B (0)$ of the hollow one?

 

[Delta2]

I think its not completely correct, we can view the field as a superposition of the fields of hollow cylinders of radius r with surface current density $\rho\omega_0\alpha$, so I believe we should consider the integral $$\mu_0\int_0^{\alpha} \rho \omega_0\alpha dr$$ and that should give the field right in the center. for the field at radial position ${r_0}$ we have to find the integral $$\mu_0\int_{r_0}^{\alpha} \rho \omega_0\alpha dr$$ cause only the outer hollow cylinders will contribute.

 

[Delta2]

So the left hand side of Ampere's law ( the path integral) is equal to $(B(0,0)+B(r,0))L$. What is the right hand side equal to?

I believe I have a sign error here it should be$$(B(0,0)-B(r,0))L$$

So i believe we now end up with the equation of post #28. Setting the boundary condition that $B(\alpha)=0$ (field vanishes at $r=a$) you can calculate $B(0)=\mu_0\rho\omega_0\alpha^2$ from that equation at post #28.

 

[Delta2]

What is the solution provided by your teacher, is it different from $$B(r)=\mu_0\rho\omega_0\alpha(\alpha-r)\hat z$$. The approach you described (with a different amperian loop) might work as well but I don't think it gives a different result.

 

[sagigever]

What is the solution provided by your teacher, is it different from $$B(r)=\mu_0\rho\omega_0\alpha(\alpha-r)\hat z$$. The approach you described (with a different amperian loop) might work as well but I don't think it gives a different result.

Yes that's the answer but than I realized you said the same just different aproach, although I still not feel comfortable with the method you suggested but I guess I will have to work about it myself, btw can you help me understand physicaly why in such case of symmetry the magnetic field outside is zero? I understand the math but if you can provide me physical explantion?

 

[Delta2]

I don't have a very good physical/intuitive explanation for that. My explanation involves taking advantage of the cylindrical symmetry, using Ampere's law and taking the proper amperian loop and doing the math we can conclude that the field outside will be zero. Ok there is a crucial detail here that the total current enclosed by the loop will be zero because the $\hat\phi$ current will be canceled by the diametrical $-\hat\phi$ current. If you want I can post details. The amperian loop will be a rectangular with both of each sides (the sides parallel to the axis of the cylinder) outside the cylinder.

 

[sagigever]

I don't have a very good physical/intuitive explanation for that. My explanation involves taking advantage of the cylindrical symmetry, using Ampere's law and taking the proper amperian loop and doing the math we can conclude that the field outside will be zero. Ok there is a crucial detail here that the total current enclosed by the loop will be zero because the $\hat\phi$ current will be canceled by the diametrical $-\hat\phi$ current. If you want I can post details. The amperian loop will be a rectangular with both of each sides (the sides parallel to the axis of the cylinder) outside the cylinder.

But I can say it on every symmetric loop, even if it in the cylinder, if I understood you correctly