Homework Help: Magnetic Field of half cylinder

1. Jul 31, 2011

n0va

1. The problem statement, all variables and given/known data
Electrical current is flowing along a long, straight wire that has a shape of a half-cylinder of radius R. Cross-section of the wire is shown, current is directed out of the plane of the picture. Current is distributed uniformly along the wire and has total magnitude I. Find magnetic field B at the point O on axis of the wire.

2. Relevant equations
dB = $\mu$0/4pi (IdlXr)/r^2

3. The attempt at a solution

Here's what i don't get? is this saying that the wire is cut in half and is only a half cylinder and current is coming out? or is the wire bent in a half cylinder and is asking for the mag field at a distance R? btw the point O on the axis is literally the center of the circle. From O to to anywhere on the semicircle is the radius R. I'm confused as to the meaning.

2. Jul 31, 2011

kuruman

The wire is not bent. The problem clearly states "straight wire". It is a wire cut in half along its length so that its cross section is a semicircle.

3. Jul 31, 2011

n0va

okay so when I find the magnetic field due to a small length dL, do i just integrate that over the course of the half circle piR or do i also include the radius? because if the wire is just a semi circle, then does the straight line segment also carry a charge?

4. Jul 31, 2011

n0va

actually can i even use the biot savart law in this case? because the law is used for infinitesimal current elements only and in this case, the distance between the current element and point of interest is very small seeing that the point of interest is on the wire itself...

5. Jul 31, 2011

kuruman

Don't use Biot-Savart - it will be too messy. Calculate the contribution dB from an infinite wire of cross sectional area (rdθ)xdr and then do the double integral over the semicircle.

6. Jul 31, 2011

n0va

so does this make sense?

dL = Rdθ

area of the cross sectional area with arc length dl = area = $\frac{dl}{2R\pi}$$\pi$R2 = RdθR2$\pi$/2R$\pi$ = R2dθ/2

so do i do that cross r? and why are we using the area in this case?

7. Jul 31, 2011

kuruman

We are using the area because we know that the current density J is the total current I carried by the wire divided by the area over which it flows.

Here is a step by step procedure:
1. Find the current density.
2. Find the current dI carried by an elemental wire of cross sectional area (rdθ)xdr.
3. Find the magnitude of the magnetic field dB contributed by this elemental wire at the axis using the equation for the infinite wire.
4. Find the x and y components of the magnetic field, x and dBy.
5. Integrate each component separately.

If you can do all this, fine. If not, please post a detailed account of what you did and/or where you got stuck.