Magnetic field in a rotating uniformly-charged infinite cylinder

[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

 

[Delta2]

Yes you can do that, if you carefully apply Ampere's law for a symmetrical rectangular loop inside the cylinder you ll end up with an equation of 0(path integral)=0(total current enclosed).

Now that I think of it , you get 0=0 even if the symmetrical loop has its two parallel to the axis of the cylinder, sides outside the cylinder. So sorry it seems that my proof at post #34 doesn't work.

 

[Delta2]

The only argument I see now is based on symmetry. Due to cylindrical symmetry we can argue that the field outside the cylinder will be in the z-direction. But again due to symmetry we can't decide if it will point in the up or down direction. So the only symmetrical option is to be zero.