Magnetic Field of half cylinder

In summary, the student was trying to solve an equation for the magnetic field at a point on an infinite wire, and was getting stuck on Step 4. They posted a step by step procedure to help others follow along.
  • #1
n0va
7
0

Homework Statement


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.


Homework Equations


dB = [itex]\mu[/itex]0/4pi (IdlXr)/r^2


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.
 
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  • #2
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
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
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
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
kuruman said:
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.

so does this make sense?

dL = Rdθ

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

so do i do that cross r? and why are we using the area in this case?
 
  • #7
n0va said:
so does this make sense?

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

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

FAQ: Magnetic Field of half cylinder

What is the magnetic field of a half cylinder?

The magnetic field of a half cylinder depends on the distance from the center of the cylinder, the radius of the cylinder, and the strength of the current flowing through it. It also varies depending on whether the cylinder is hollow or solid.

How is the magnetic field calculated for a half cylinder?

The magnetic field of a half cylinder can be calculated using the Biot-Savart law, which takes into account the current, distance, and geometry of the cylinder. Alternatively, it can be calculated using the Ampere's law for a current loop.

What factors affect the strength of the magnetic field in a half cylinder?

The strength of the magnetic field in a half cylinder is affected by the current flowing through it, the distance from the center of the cylinder, and the radius of the cylinder. Additionally, the magnetic permeability of the material used for the cylinder can also impact the strength of the field.

How does the direction of the magnetic field change in a half cylinder?

The direction of the magnetic field in a half cylinder changes as you move around the cylinder. At the center, the field is perpendicular to the surface of the cylinder, while at the edges it is parallel to the surface. Additionally, the direction of the field can also be affected by the direction of the current flow.

What are the applications of the magnetic field of a half cylinder?

The magnetic field of a half cylinder has various applications in engineering and technology. It is used in the design of electric motors, generators, and transformers. It is also utilized in medical imaging techniques such as magnetic resonance imaging (MRI). Additionally, it is used in magnetic levitation systems and particle accelerators.

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