Calculating Current in a Cylindrical Region

[Aviegaille]

Homework Statement

(a)The current density across a cylindrical region of radius R varies according to the equation: J=J0(1-r/R), where r is the distance from the axis of the cylinder. The current density is the maximum J0 at the axis r=0 and decreases linearly to zero at the surface r=R. Calculate the current in terms J0 and the region's cross sectional area A=pi*R^2.

(b) Now suppose that a current density was a maximum Jo at the surface and decreased linearly to zero at the axis, so that: J=J0 r/R. Calculate the current. Why is the result different for these two cases?

Homework Equations

I=JA

The Attempt at a Solution

I uploaded a picture of the first part but I am not sure if it's correct. I also don't know how to get the area from this problem. I am thinking of plugging the value of R from I to get the area but I am pretty sure it is not right.

 

[Simon Bridge]

You cannot just use the total cross-section because the current density is different in different parts of the cylinder.
Instead you have to add up the contributions from each small part of the area.
i.e. you need to set up an integral.

If I is the current and J is the current density, then dI = J.dA

 

[Aviegaille]

You cannot just use the total cross-section because the current density is different in different parts of the cylinder.
Instead you have to add up the contributions from each small part of the area.
i.e. you need to set up an integral.

If I is the current and J is the current density, then dI = J.dA

Can you elaborate how can I use that equation?

 

[Simon Bridge]

You integrate both sides.
You need an expression for J in terms of r and an expression for dA in terms of dr.
Hint: how much current passes through the area between r and r+dr?

 

[HalfLostAndSearching]

As a scientist, it is important to always double check your work and make sure that your assumptions and calculations are correct. In this case, your attempt at a solution for the first part appears to be incorrect. The equation I=JA is a general equation for calculating current, where I is the current, J is the current density, and A is the cross-sectional area. However, in this problem, we are given a specific equation for the current density, J=J0(1-r/R), which means that we cannot simply use the general equation without making some adjustments.

To calculate the current in terms of J0 and the region's cross-sectional area, we need to integrate the given equation for current density over the cross-sectional area of the cylinder. This can be done using the following equation:

I = ∫JdA

where the integral is taken over the entire cross-sectional area of the cylinder. In this case, the cross-sectional area is a circle with radius R, so the integral becomes:

I = ∫J(r)2πrdr

where J(r) is the current density at a distance r from the axis of the cylinder. Plugging in the given equation for J(r), we get:

I = ∫J02πr(1-r/R)dr

Evaluating this integral, we get:

I = J0πR2/4

So, the current in terms of J0 and the region's cross-sectional area is:

I = J0πR2/4

For the second part of the problem, we are given a different equation for the current density, J=J0 r/R. This means that the current density is now maximum at the surface of the cylinder and decreases linearly to zero at the axis. To calculate the current in this case, we can use the same approach as before, but with a different equation for J(r):

I = ∫J(r)2πrdr = J0πR2/2

So, the current in this case is:

I = J0πR2/2

The result is different for these two cases because the current density is distributed differently in the two cases. In the first case, the current density is maximum at the axis and decreases linearly to zero at the surface, whereas in the second case, the current density is maximum at the surface and decreases linearly to zero at the axis. This difference in distribution leads to