# Ampere's Law: Cylindrical conducter with varying current

• Renaldo
In summary, according to the homework statement, the current density in the region from zero to R varies as J(r) = J0e−r/R. The magnitude of the magnetic field in the regions r < R and r > R is μ0J0e−r/R(∏r2).
Renaldo

## Homework Statement

The current density of a cylindrical conductor of radius R varies as J(r) = J0e−r/R (in the region from zero to R). Express the magnitude of the magnetic field in the regions r < R and r > R. (Use any variable or symbol stated above along with the following as necessary: μ0.)

Produce a sketch of the radial dependence, B(r).

## Homework Equations

Ampere's Law

$\oint B \bullet ds$ = μ0ienc

## The Attempt at a Solution

At r < R:

B = μ0J(r)A/2∏r

J(r) = J0e−r/R
J(r)A = J0e−r/R(∏r2)

B = (μ0J0e−r/Rr)/2

This is not correct.

How current is defined in terms of current density?

(Hint: The relation between current and current density involves a integral.)

I = J(r)A
dI = J0∏r2e-r/Rdr

I = J0∏∫r2e-r/Rdr

I used a computer to solve the integral, which is an integration by parts.

I = -J0∏R(r2+2rR+2R2)/er/R

B = μ0I/2∏r

B = -μ0J0R(r2+2rR+2R2)/2rer/R

Renaldo said:
I = J(r)A

You can't write that. According to the definition,
$$I=\int \vec{j}\cdot\vec{dA}$$

What is dA?

In that definition, is J constant? It seems to me that it would not be, but I don't know how to integrate

∫J0e-r/Rda

Renaldo said:
In that definition, is J constant? It seems to me that it would not be, but I don't know how to integrate

∫J0e-r/Rda

No, J is not constant. You have the expression for J and it varies with r. About the differential area dA, see the attachment. We have to find the total current enclosed within the Amperian loop. As we move out away from the axis of cylinder, the current density varies so we select a very small area (differential area, dA) where we can assume that current density is effectively constant. Then we sum up (or integrate) the expression we get for the current passing through that small region. See the attachment. Can you calculate dA now? I have drawn two circles of radius r and the other with r+dr. The shaded region is dA.

#### Attachments

• cylinder.png
7 KB · Views: 440
da = 2∏rdr

2∏J0∫e(-r/R)rdr = -2J0∏R(r+R)/er/R

B = -μ0J0R(r+R)/rer/R

That's what I get and it isn't correct.

Last edited:
What were your limits? Do you know how to evaluate a definite integral?

limits of integration were from 0 to r, r < R.

Renaldo said:
limits of integration were from 0 to r, r < R.

Right but check your work again. I get a different answer. (##e^0=1##)

## What is Ampere's Law?

Ampere's Law is a physical law in electromagnetism that relates the magnetic field produced by a closed loop of current to the current passing through that loop.

## What is a cylindrical conductor?

A cylindrical conductor is a conductor that has a circular cross section and is in the shape of a cylinder. It is commonly used in electrical circuits and can come in various sizes and materials.

## What does it mean for a cylindrical conductor to have varying current?

When a cylindrical conductor has varying current, it means that the current passing through it is changing or fluctuating over time. This can be due to various factors such as the circuit design or external influences.

## How does Ampere's Law apply to a cylindrical conductor with varying current?

Ampere's Law states that the magnetic field produced by a closed loop of current is directly proportional to the current passing through that loop. Therefore, in a cylindrical conductor with varying current, the magnetic field will also vary and change accordingly.

## What are some practical applications of Ampere's Law in relation to cylindrical conductors with varying current?

Ampere's Law has many practical applications, such as in the design of electromagnets, motors, and generators. It is also used in calculating and predicting the behavior of magnetic fields in various scenarios, such as in MRI machines and particle accelerators.

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