Steady Current and Magnetic Field

In summary, the problem asks for the magnetic field within the smaller of two current-carrying cylinders. By symmetry, the field is e_theta. The x component of the magnetic field should cancel when properly added.
  • #1
SimbaTheLion
8
0

Homework Statement



Question 3 of http://www.damtp.cam.ac.uk/user/examples/B10b.pdf .

Homework Equations



[itex]\nabla \times B = \mu_0 J[/itex]

The Attempt at a Solution



I know for a cylinder, J (vector) = J (scalar) * k (vector), where unit k is the vector in the z-direction. So [itex]\nabla \times B = (0, 0, \mu_0 J)[/itex], somewhere (not quite sure where this is true in terms of a, b, d). I think I need to find B for the big cylinder, then B for the small cylinder, and combine them somehow? I don't see why the field in between the cylinders is important; we're only interested in B for x² + y² < a².

Thanks!
 
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  • #2
I think you have everything you need to do this problem. Write down explicitly the sum of the two fields, one from +J and one from -J, for any point in the current free region of interest. The x dependence should drop out.
 
  • #3
Spinnor said:
I think you have everything you need to do this problem. Write down explicitly the sum of the two fields, one from +J and one from -J, for any point in the current free region of interest. The x dependence should drop out.

If I was not clear enough, the problem suggested what to do.
 
  • #4
Spinnor said:
I think you have everything you need to do this problem. Write down explicitly the sum of the two fields, one from +J and one from -J, for any point in the current free region of interest. The x dependence should drop out.

What is the 'free region of interest'? If I take a +J thing from one cylinder and then subtract the J bit from the other cylinder, surely I'm going to end up with an expression for the stuff in between the two cylinders? The question asks for the field within the smaller cylinder, not between the two.

My notes on a vaguely similar problem claim B = B(r)e_theta 'by symmetry'. Not quite sure why it's e_theta by symmetry.
 
  • #5
SimbaTheLion said:
What is the 'free region of interest'? If I take a +J thing from one cylinder and then subtract the J bit from the other cylinder, surely I'm going to end up with an expression for the stuff in between the two cylinders? The question asks for the field within the smaller cylinder, not between the two.

My notes on a vaguely similar problem claim B = B(r)e_theta 'by symmetry'. Not quite sure why it's e_theta by symmetry.

You break the complex problem down into two simple ones. If you had a single cylinder with current density J you could tell me the magnetic field for any radius r inside the conductor via Ampere's law. To this field you add the imaginary field from the current -J which flows in the smaller cylinder. The sum of J and -J = 0, flows in the small cylinder. Properly added the x component of the magnetic fields should cancel.

If you are still stuck I'll try and add more.
 

1. What is a steady current?

A steady current is a continuous flow of electric charge, typically in the form of electrons, through a conductor. It is also known as direct current (DC) and is characterized by a constant magnitude and direction.

2. How is a magnetic field created by a steady current?

A magnetic field is created by a steady current due to the movement of electric charges. As the electrons flow through a conductor, they create a circular magnetic field around the conductor, with the direction of the field determined by the direction of the current.

3. What is the relationship between the strength of a steady current and the strength of the magnetic field it creates?

The strength of a magnetic field created by a steady current is directly proportional to the strength of the current. This means that as the current increases, the magnetic field also increases in strength.

4. How does the direction of the current affect the direction of the magnetic field?

The direction of the magnetic field created by a steady current is always perpendicular to the direction of the current. This means that if the current is flowing in a straight line, the magnetic field will form circles around the conductor in a perpendicular direction.

5. How can a steady current and magnetic field be used in practical applications?

Steady currents and magnetic fields have a wide range of practical applications. Some examples include electromagnets used in motors and generators, magnetic levitation trains, and magnetic resonance imaging (MRI) machines used in medical diagnostics.

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