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I have a simple question. I need to find out what is the magnetic field inside a rotating long cylinder with a charged density $\sigma$. The cylinder rotates around its axis with angular speed $\omega$. All of the parameters are constants.

I've tried using Amper's law but I can't seem to get to the right answer.

$J=\lambda \cdot v=\sigma\omega R dr$

$I=\sigma \omega R L$

L is just the the part of the length of the whole cylinder. So I i'll just use Amper's law :

$\mu_0 \cdot I_{in}=B\cdot 2\cdot\pi \cdot L $

$\mu_o\sigma \omega R \cdot2L=B\pi 2L$

$B=\mu_o \sigma \omega R/\pi$

The loop is a circle with a radius of $L$, and so the current flows thorough a line of length $2L$.

The correct answer seems to be the same answer just with the division by $\pi$.

Where did I go wrong?

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Your choice of Amperian loop is inappropriate for this situation. In these types of problems, you need to choose an Amperian loop where the length of the loop is either parallel or perpendicular to the field that you expect to arise from the situation (try to, its not always possible to choose something that has a convenient shape, but the symmetry of this situation allows it). The real equation for ampere's law is:

$$\oint_L \vec{B}\cdot d\vec{l} =\mu_0 I_{enc}$$

where $L$ is the path of your Amperian loop. The dot product here signifies that only the component of $\vec{B}$ along the path contributes, so choosing a path along the field lines will simplify things greatly (or if its perpendicular, it contributes zero).

Given the axial symmetry of the case you have here, you expect the field lines inside of the cylinder to point along the axis of the cylinder, much like in the case of the long solenoid. Therefore, it would be better to choose a square Amperian loop, much like that of the linked example in the last sentence.

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  • $\begingroup$ Thanks for telling me where I went wrong. I built a rectangular loop instead of the circular. Two of the loop's edges are perpendicular to the magnetic field (which is in the direction of the axis) hence, their contribution to the field is zero, and the two remaining edges are parallel to the field's direction. So the integral become very simple. $\endgroup$ – Alexs68 Dec 21 '15 at 20:22
  • $\begingroup$ No problem. I'm glad that it worked out. If you can take advantage of the symmetry arguments it can save you a lot of work on these types of problems. So this is good practice. $\endgroup$ – tmwilson26 Dec 21 '15 at 20:25

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