A long, straight wire with a circular cross section of radius R carries a current I. Assume that the current density is not constant across the cross section of the wire, but rather varies as J=αr, where α is a constant.

A) By the requirement that J integrated over the cross section of the wire gives the total current I, calculate the constant α in terms of I and R.
Express your answer in terms of the variables I, R, and appropriate constant (π).
B) Use Ampere's law to calculate the magnetic field B(r) for r≤R. Express your answers in terms of I.
Express your answer in terms of the variables I, R, r, and appropriate constants (μ0 and π).
C) Use Ampere's law to calculate the magnetic field B(r) for r≥R. Express your answers in terms of I.
Express your answer in terms of the variables I, R, r, and appropriate constants (μ0 and π).

I was able to figure out A) but I don't know how to use that answer for B) and C)

To calculate the magnetic field for B(r) when r ≤ R (B(r) inside the wire) and r ≥ R (B(r) outside the wire), we can use Ampere's Law. Ampere's Law relates the magnetic field around a closed loop to the current passing through the loop.

A) In order to calculate α in terms of I and R, we know that the current density J integrated over the cross section of the wire should give the total current I. The current density J is given by J = αr.

To integrate J over the cross section of the wire, which is a circle of radius R, we use the formula for the area of a circle: A = πR^2. The integral of J over the cross-sectional area is then:

I = ∫J dA
= ∫(αr) dA
= α ∫r dA
= α ∫r d(πR^2)
= α ∫r πR^2 dr

To evaluate this integral, we need to determine the limits of integration. Since the wire is long and straight, we consider a cross-sectional slice at distance r from the axis of the wire. The limits of integration would be from r = 0 (center of the wire) to r = R (outer edge of the wire). Therefore, the integral becomes:

I = α ∫r πR^2 dr, integrated from r = 0 to r = R.

Integrating, we have:

I = α πR^2 ∫r dr, integrated from r = 0 to r = R.

The integral of r with respect to r is (1/2)r^2, so:

I = α πR^2 [ (1/2)R^2 - (1/2)(0)^2 ]
I = α πR^2 (1/2)R^2
I = (α π/2)R^4

Solving for α, we have:

α = (2I)/(πR^4)

Therefore, the constant α in terms of I and R is α = (2I)/(πR^4).

Now, let's move on to B) and C) to calculate the magnetic field B(r) for both cases.

B) Inside the wire (r ≤ R, B(r) inside):
For r ≤ R, we need to find the magnetic field B(r) using Ampere's Law. According to Ampere's Law, the line integral of the magnetic field B(r) around a closed loop is equal to the product of the permeability of free space (μ0) and the total current passing through the loop.

Consider a circular loop of radius r (r ≤ R) within the wire. The magnetic field B(r) is constant along this loop, and its magnitude must be the same at every point on the loop. Using Ampere's Law, we have:

∮ B(r) · dl = μ0I_enclosed

Since B(r) is constant along the loop, we can take it out of the line integral:

B(r) ∮ dl = μ0I_enclosed

The line integral ∮ dl is equal to the circumference of the circular loop, which is 2πr. Therefore, we have:

B(r) (2πr) = μ0I_enclosed

Since the current enclosed by the circular loop is the total current I, we can write:

B(r) (2πr) = μ0I

Solving for B(r), we have:

B(r) = (μ0I) / (2πr)

We express our answer in terms of the variables I, R, r, and the appropriate constant μ0 and π.

C) Outside the wire (r ≥ R, B(r) outside):
For r ≥ R, we use Ampere's Law again. Consider a circular loop of radius r (r ≥ R) outside the wire. Since there is no current enclosed by the loop outside the wire, the line integral ∮ B(r) · dl must be zero. Therefore:

∮ B(r) · dl = 0

Since B(r) is constant along the loop, we can take it out of the line integral:

B(r) ∮ dl = 0

The line integral ∮ dl is equal to the circumference of the circular loop, which is 2πr. Therefore:

B(r) (2πr) = 0

This equation tells us that the magnetic field B(r) outside the wire is zero (B(r) = 0) for r ≥ R.