Magnetic field within cylinder using Ampere's Law

In summary: The circulation of B-field around this rectangle will be equal to the current going through the rectangle.In summary, using Ampere's law, it is shown that the magnetic field strength in a region within a cylinder with constant current density is equal to (mu-nought)*j*r/2. The field is in the theta-hat direction and there is no component in the r-hat direction. To prove that there is no component in the z-hat direction, a rectangle can be used to show that the circulation of the B-field is equal to the current going through the rectangle.
  • #1
Silversonic
130
1

Homework Statement



Using Ampere's law, show that the magnetic field strength in a region within a cylinder, which has a constant current density j (flowing in the direction parallel to its axis), is equal to

B = (mu-nought)*j*r/2

The Attempt at a Solution



It doesn't say specifically, but this is the field in the theta-hat direction - i.e. in the direction of the cylinder's axis of rotation. I can prove this easily and the actual question isn't the problem. I'm assuming that this question means I have to show that the magnetic field in the r-hat direction (radially outwards) and the z-hat direction (in the direction of the current flow) are both zero.

I can prove there is no component in the r-hat direction by taking an imaginary cylinder, placing it within and using the fact that the flux through the cylinders surface is always equal to zero.

However, how do I prove that there is no component in the z-hat direction? Any help/hints appreciated.
 
Physics news on Phys.org
  • #2
Silversonic said:
I can prove there is no component in the r-hat direction by taking an imaginary cylinder, placing it within and using the fact that the flux through the cylinders surface is always equal to zero.

However, how do I prove that there is no component in the z-hat direction? Any help/hints appreciated.

Did you try using the same reasoning? That is, finding a surface that would allow to calculate the circulation of z - component of the field?
 
  • #3
Inna said:
Did you try using the same reasoning? That is, finding a surface that would allow to calculate the circulation of z - component of the field?

What surface would I use? If I use a cylinder, that only tells me that the flux in the z-hat direction through one end is equal and opposite to the flux in the z-hat direction flowing through the other end - but it doesn't tell me that the magnetic field for that component is zero, only that the sum of the two fluxes flowing through the ends sums to zero.
 
  • #4
Wait - it seems like you are using a Gauss law instead of Ampere's law. You need a circulation of B-field around the boundary of your surface. It will be proportional to the current going through the surface.
 
  • #5
Inna said:
Wait - it seems like you are using a Gauss law instead of Ampere's law. You need a circulation of B-field around the boundary of your surface. It will be proportional to the current going through the surface.

I used Ampere's law to find the direction of the field in the theta-hat direction, I haven't touched Gauss' law at all.

I used the fact that [tex]\int[/tex]B.dS = 0, using a closed cylinder, to show that it is zero in the r-hat direction, but can't prove it for the z-hat direction.
 
  • #6
Try a rectangle with one side placed inside the cylinder, parallel to the z-axis, and the opposite side outside of the cylinder.
 

1. What is Ampere's Law and how does it relate to magnetic fields within a cylinder?

Ampere's Law is a mathematical equation that describes the relationship between electric currents and magnetic fields. It states that the magnetic field around a closed loop is proportional to the electric current passing through that loop. This law can be used to calculate the magnetic field within a cylinder by integrating the current density over the cross-sectional area of the cylinder.

2. What is the formula for calculating the magnetic field within a cylinder using Ampere's Law?

The formula for calculating the magnetic field within a cylinder using Ampere's Law is B = μ0I/2πr, where B is the magnetic field, μ0 is the permeability of free space, I is the current passing through the cylinder, and r is the distance from the center of the cylinder.

3. How does the direction of the magnetic field within a cylinder vary with the direction of the electric current?

The direction of the magnetic field within a cylinder is dependent on the direction of the electric current passing through it. According to the right-hand rule, if the current is flowing in a clockwise direction, the magnetic field will be directed outwards from the center of the cylinder, and if the current is flowing in a counterclockwise direction, the magnetic field will be directed inwards towards the center of the cylinder.

4. What factors can affect the magnetic field within a cylinder using Ampere's Law?

The magnetic field within a cylinder using Ampere's Law can be affected by several factors, including the magnitude and direction of the electric current, the distance from the center of the cylinder, and the permeability of the material inside the cylinder. Additionally, any changes in the current density or shape of the cylinder can also impact the magnetic field.

5. How is the magnetic field within a cylinder using Ampere's Law different from the magnetic field around a straight wire?

The magnetic field within a cylinder using Ampere's Law is similar to the magnetic field around a straight wire, as both are described by the same mathematical equation. However, in the case of a cylinder, the magnetic field is more uniform and circulates around the entire cross-sectional area of the cylinder, while in a straight wire, the magnetic field is strongest near the wire and decreases as you move away from it.

Similar threads

  • Introductory Physics Homework Help
Replies
12
Views
194
  • Introductory Physics Homework Help
Replies
2
Views
182
  • Introductory Physics Homework Help
Replies
7
Views
983
  • Introductory Physics Homework Help
Replies
25
Views
270
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
16
Views
388
  • Introductory Physics Homework Help
Replies
2
Views
64
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
18
Views
1K
Back
Top