Ampére's Law Cylindrical Conductor with Hole

In summary, a long circular rod made of conducting material has a cylindrical hole bored parallel to its axis and displaced from the center by a distance d. The rod carries a current I distributed uniformly over its cross-section. By considering the superposition of two currents flowing in opposite directions, it can be shown that the magnetic field in the hole is equal to \frac{\mu_{0} dI}{2 \pi (R^{2} - a^{2})}. Using Ampere's law, the direction of the magnetic field can be determined and the vector sum of the magnetic fields due to the two rods can be calculated to find the final magnetic field in the empty region.
  • #1
Frinkz
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Homework Statement


A long circular rod of radius R, made of conducting material, has a cylindrical hole of radius a bored parallel to its axis and displaced from the centre of the rod by a distance d. The rod carries a current I distributed uniformly over its cross-section.

Consider the superposition of two currents flowing in opposite directions, and hence show that the magnetic field in the hole is uniform and equal to:

[tex]B = \frac{ \mu_{0} dI }{2 \pi (R^{2} - a^{2})}[/tex]

Hint:
Consider the rod with a hole to be a superposition of two rods without holes: one with current density j and one with current density -j (the latter representing the hole)

Deduce from Ampére's law that:
[tex]\mathbf{B}(\mathbf{r})=\frac{1}{2} \mu_{0}(\mathbf{j} \times \mathbf{r})[/tex]
for each current, where j is the current density, r is the vector from centre of the rod.


Homework Equations


Ampéres law (integral form):
[tex]\oint_ L \mathbf{B} \cdot d\mathbf{L} = \mu_{0} I[/tex]

(differential form)
[tex]\mathbf{\nabla} \times \mathbf{B} = \mu_{0} \mathbf{j}[/tex]

The Attempt at a Solution


Using cylindrical polars, I worked through for the integral form of Ampere's law for the first rod (j, radius R) and got:
[tex]B = \frac{\mu_{0}}{2} j[/tex]
I think that's right, and it seems to be along the lines of what I should get. Except, it's a scalar, but what I'm given in the hints is vector.

Also, given that r is the vector from the centre of the rod to axis is going to be zero for the first rod, doesn't j x 0 give zero?

But if I try the differential form I end up with zero after the first step, I think this isn't the way to do it?

So I'm a bit stuck, any hints would be appreciated :)
 
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  • #2
Frinkz said:
Using cylindrical polars, I worked through for the integral form of Ampere's law for the first rod (j, radius R) and got:
[tex]B = \frac{\mu_{0}}{2} j[/tex]

You can guess the direction of the field at each point from the Biot-Savart law (all the [tex] dl \times r[/tex] bits add up).

When you apply Ampere's law, you'll need to draw a loop so that you can evaluate the line integral, as well as calculating the current which is enclosed by that loop. You'll be able to evaluate the magnitutde of B this way for the rod with +I and the rod with -I. Since you've used [itex]\oint_ L \mathbf{B} \cdot d\mathbf{L}[/itex] as [itex]2 \pi r B[/itex] in your calculation, you've already figured out the direction of B.

Note that the two wires of current +I and -I will have different current densities as their areas are different. Finally, find the vector sum of the magnetic field due to the contribution of the two rods to get your final magnetic field in the empty region.
 
Last edited:

What is Ampére's Law?

Ampére's Law is a fundamental law in electromagnetism that describes the relationship between electric currents and the magnetic field they produce.

What is a cylindrical conductor with a hole?

A cylindrical conductor with a hole is a type of conductor with a cylindrical shape and a hole in the center. It is commonly used in electrical engineering and physics experiments to study the effects of electric currents on magnetic fields.

How does Ampére's Law apply to a cylindrical conductor with a hole?

Ampére's Law states that the line integral of the magnetic field around a closed loop is equal to the current passing through the loop multiplied by a constant known as the permeability of free space. In a cylindrical conductor with a hole, this law can be used to calculate the magnetic field at any point around the hole.

What factors affect the magnetic field in a cylindrical conductor with a hole?

The magnetic field in a cylindrical conductor with a hole is affected by several factors, including the current passing through the conductor, the distance from the hole, and the orientation of the conductor. Additionally, the shape and size of the hole can also impact the magnetic field.

How is Ampére's Law used in practical applications?

Ampére's Law is used in various practical applications, such as designing electromagnets, calculating the magnetic field in electric motors, and studying the behavior of magnetic materials. It is also used in medical imaging techniques, such as magnetic resonance imaging (MRI).

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