Finding the Auxillary Field of a Long Copper Rod

[thatguy14]

Homework Statement

I am looking at an example problem in the text and they skipped some steps. I think I am missing somthing obvious but none-the-less I don't know what is going on.

We have a long copper rod o radius R which carries a uniformly distributed (free) current I. Find H, the auxillary field inside and outside the rod.

Homework Equations

$\oint H dot dl$ = Free current enclosed

The Attempt at a Solution

so the LHS would be (magnitude only) H*2pis for some amperian loop. Now I don't know how to get the Free current enclosed inside and outside the rod.

Thanks for your help!

 

[Stealth95]

The Amperian loops are always closed curves (that's because the line integral at LHS is closed), so they are always the circumference of a surface. In our case this is the surface of a circle with radius $\displaystyle{s}$. The current at RHS is the total current that goes through this surface. When you are outside of the cylinder it's easy to find this current. When you are inside, you just have to take into account that the current is uniformly distributed.

 

[thatguy14]

I mostly understand that. The answer the book gives for I free is = (I*pi*s^2)/(pi*R^2) which makes sense, but I am looking for what definition they used exactly to get that. What equation?

 

[Stealth95]

You can get this by using the current density vector $\displaystyle{\vec{J}}$. This is the current that flows through a unit area of the wire crossection (or for uniform distributions the ratio current/area). For this case the current flows only in a specific direction, so it's not necessary to use vectors. We can just say that:
$$\displaystyle{J=\frac{I}{\pi R^2}}$$
because the current is uniformly distributed. The current we need is given by a surface integral:
$$\displaystyle{I_{f,enc}=\int _{S}JdS}$$
where $\displaystyle{S}$ is the circle's surface. Actually the integral is an overkill, because $\displaystyle{J}$ is constant here. If you have seen any applications of Gauss's law, it's like using the charge density in electrostatics.

 

[paranormal]

As a scientist, it is important to thoroughly understand the problem and all the necessary equations before attempting to solve it. In this case, the key equation is the Ampere's law, which relates the magnetic field (H) to the current enclosed within a closed loop. In order to find the auxiliary field inside and outside the rod, we need to consider the current distribution within the rod.

The first step would be to draw a diagram of the copper rod and its current distribution. Since the current is uniformly distributed, we can assume that it is flowing in a straight line along the length of the rod. This means that the current enclosed within any closed loop surrounding the rod would be equal to the total current flowing through the rod.

To find the auxiliary field inside the rod, we can choose an amperian loop that lies completely within the rod. This means that the current enclosed would be equal to the total current flowing through the rod. Plugging this into Ampere's law, we can solve for H.

To find the auxiliary field outside the rod, we can choose an amperian loop that surrounds the rod. In this case, the current enclosed would include the current flowing through the rod as well as the current flowing outside the rod. Again, using Ampere's law, we can solve for H.

It is important to note that the auxiliary field is only dependent on the current enclosed within the amperian loop, not the shape or size of the loop. This means that we can choose any convenient loop to solve for H, as long as it encloses the relevant current.

In summary, the key to finding the auxiliary field in this problem is understanding the current distribution and selecting appropriate amperian loops to apply Ampere's law. I hope this helps clarify the steps that were skipped in the example problem. If you are still having trouble, I suggest consulting your textbook or seeking help from a classmate or professor. Good luck!