# Magnetic field of long cylindrical wire

i have been studying electromagnetism on my own and I got a question about an exercise in David Griffiths, Introduction to Electrodynamics book.

So I'm asked to calculate the magnetic field both inside and outside the wire created by a steady current $$I$$ that flows down a long cylindrical wire of radius $$a$$ in the $$x$$ direction.

I know the solution but I don't understand it and was hoping you guys could help me out, By the way we can use the Ampere's Law. One of my confusions is that the direction is going to be circumferential and I don't really understand what this means, also I don't get why it is going to be zero when $$s$$ is lower than $$a$$. Any help is appreciated.

Thanks.

In (a) you can understand how the direction of the magnetic flux is formed by using the symmetry of the uniformly distributed current. On each point you can be sure that the direction is circumferential because the contribution of every current flow to the magnetic field above and below this point along the wire are compensating each other. As a result you only have a component "to the side". The magnetic field is circling around the wire and becomes weaker in radial direction. Here is a picture as an example: To understand why $$B(r)=0$$ for $$r you can use Ampere's law: $$\begin{equation} \oint\vec{B}*d\vec{s}=\mu_0*I \end{equation}$$ You only have a flowing current on the outside surface, so inside the integration curve (for example circle at $$r) $$I=0$$. As a result $$\vec{B}=0$$. In (b) you can use the same argument for the direction of $$\vec{B}$$ as in (a) but here you get a non zero result for the magnetic field inside the wire using Ampere's law. I hope I could help you.