Ampere's Law: Constant Magnetic Field b/w Non-Coaxial Cylinders

[Merrank]

The question is:

Between two long parallel cylinders of radius "a" and "b" (non-coaxial) and an axal separation of "c", a steady current of "I" flows. (See attachment below) Show that the inner cylinder (radius "a") has a constant magnetic field. Use Ampere's Law. Indicate all steps clearly. [Hint: 0 = 1 + (-1)]

 

[Andrew Mason]

The question is:

Between two long parallel cylinders of radius "a" and "b" (non-coaxial) and an axal separation of "c", a steady current of "I" flows. (See attachment below) Show that the inner cylinder (radius "a") has a constant magnetic field. Use Ampere's Law. Indicate all steps clearly. [Hint: 0 = 1 + (-1)]

You have to use the principle of superposition. Work out the field for the large cylinder having a uniform current density (ie. uniform over its cross-sectional area) using Ampere's law; then work out the field for the small cylinder with a current flowing in the opposite direction with a cross-sectional density that that is the same magnitude (but opposite in direction) to that of the larger cylinder. Superimpose the two (add the field vectors).

AM

 

[thepatientmental]

To show that the inner cylinder (radius "a") has a constant magnetic field, we can use Ampere's Law, which states that the line integral of the magnetic field around a closed loop is equal to the product of the enclosed current and the permeability of free space (μ0). In this case, we will consider a circular loop around the inner cylinder (radius "a") with a radius of r and a length of L, and use Ampere's Law to calculate the magnetic field at any point on the loop.

Step 1: Setting up the equation

We start by writing Ampere's Law in integral form:

∮B⃗ ⋅ dl⃗ = μ0Ienc

Where B⃗ is the magnetic field, dl⃗ is an infinitesimal element of the closed loop, and Ienc is the enclosed current. In this case, we will be considering a circular loop, so dl⃗ can be written as rdθ, where θ is the angle between the direction of the loop and the direction of the magnetic field.

Step 2: Simplifying the equation

We can simplify the equation by noting that the magnetic field is constant along the loop, and thus can be taken outside the integral. We can also substitute the value of dl⃗ with rdθ, and rewrite the enclosed current as the current density (J) multiplied by the area of the loop (πr2).

∮B⃗ ⋅ rdθ = μ0Jπr2

Step 3: Solving for the magnetic field

We can now solve for the magnetic field (B⃗) by rearranging the equation:

B⃗ = μ0Jπr2/∮rdθ

Step 4: Evaluating the integral

To evaluate the integral, we need to determine the limits of integration. Since the loop is a circle with a radius of r, the limits of integration will be from 0 to 2π, representing a full revolution around the loop. Substituting these values into the integral, we get:

B⃗ = μ0Jπr2/∮rdθ = μ0Jπr2/2π = μ0Jr/2

Step 5: Simplifying the equation

We can further simplify the equation by noting that the current density (J) can be written as the total current (