91Ó°ÊÓ

A long, straight, solid cylinder, oriented with its axis in the \(z\) -direction, carries a current whose current density is \(\overrightarrow{\boldsymbol{J}}\). The current density, although symmetric about the cylinder axis, is not constant but varies according to the relationship $$ \begin{array}{rlr} \overrightarrow{\boldsymbol{J}} & =\frac{2 I_{0}}{\pi a^{2}}\left[1-\left(\frac{r}{a}\right)^{2}\right] \hat{k} & \text { for } r \leq a \\ & =\mathbf{0} \quad & \text { for } r \geq a \end{array} $$ where \(a\) is the radius of the cylinder, \(r\) is the radial distance from the cylinder axis, and \(I_{0}\) is a constant having units of amperes. (a) Show that \(I_{0}\) is the total current passing through the entire cross section of the wire. (b) Using Ampere's law, derive an expression for the magnitude of the magnetic field \(\vec{B}\) in the region \(r \geq a\). (c) Obtain an expression for the current \(I\) contained in a circular cross section of radius \(r \leq a\) and centered at the cylinder axis. (d) Using Ampere's law, derive an expression for the magnitude of the magnetic field \(\vec{B}\) in the region \(r \leq a\). How do your results in parts (b) and (d) compare for \(r=a ?\)

Step by step solution

01

Verify \(I_0\) as Total Current

Integrate the current density \(\overrightarrow{\boldsymbol{J}}\) over the entire cross-sectional area to verify that \(I_0\) is indeed the total current. Set up the integral as follows: \[\int_{0}^{a} \int_{0}^{2\pi} \overrightarrow{\boldsymbol{J}} \cdot r d\theta dr\] where r is the radial distance from the cylinder axis and \(\theta\) is the angle.

02

Solve for \(I_0\)

Solving the integral from Step 1 will provide a value for \(I_0\). This is the verification that \(I_0\) is indeed the total current passing through the cross section.

03

Derive Expression for \(\overrightarrow{B}\) when \(r \geq a\)

Using Ampere's law in the form of \[ \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}\], where \(\overrightarrow{B}\) is the magnetic field, \(\mu_0\) is the permeability of free space, and \(I_{\text{enc}}\) is the current enclosed by the Amperian loop, derive an expression for the magnetic field \(\overrightarrow{B}\) when \(r \geq a\). At \(r \geq a\), the enclosed current \(I_{\text{enc}}\) for the Amperian loop of radius r is equal to \(I_0\).

04

Obtain Expression for Current \(I\) when \(r \leq a\)

For \(r \leq a\) obtain an expression for the current \(I\) contained in a circular cross section of radius \(r \leq a\) by integrating the current density \(J\) over the given area of the circular cross section. Set up the integral as follows: \[\int_{0}^{r} \int_{0}^{2\pi} \overrightarrow{\boldsymbol{J}} \cdot r d\theta dr.\] This gives a function for \(I\) depending on \(r\).

05

Derive Expression for \(\overrightarrow{B}\) when \(r \leq a\)

Using Ampere's law, derive a similar expression for the magnetic field for \(r \leq a\). Here the current \(I_{\text{enc}}\) that is enclosed in the Amperian loop of radius \(r\) is different from step 3 since it depends on \(r\) which is less than the radius of the cylinder \(a\). Set up Ampere's law as follows, where \(I\) is the current from Step 4: \[ \oint \vec{B} \cdot d\vec{l} = \mu_0 I.\] The integration over the Amperian loop will yield an expression for the magnetic field.

06

Compare Results for \(r=a\)

Lastly, insert \(r=a\) into the expressions derived for \(\overrightarrow{B}\) at \(r \geq a\) and \(r \leq a\) respectively. If the expressions are equal, the results are consistent and therefore valid.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnetic Field Calculation

Magnetic field calculation involves determining the influence a magnetic source has over a specified area. In this exercise, the magnetic field inside and outside a cylindrical wire is calculated using Ampere's Law.
Ampere's Law states: \[ \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}} \]where \( \vec{B} \) is the magnetic field, \( d\vec{l} \) is a length element along the path of integration, \( \mu_0 \) is the permeability of free space, and \( I_{\text{enc}} \) is the current enclosed by the path.
By applying Ampere's Law to a cylindrical path inside and outside the cylinder, you can derive expressions for the magnetic field in both regions.
Outside the cylinder (\( r \geq a \)), the entire current \( I_0 \) is enclosed, simplifying the calculation. Inside the cylinder (\( r \leq a \)), only the current within the radius \( r \) is enclosed, making the calculation more intricate.

Current Density

Current density \( \overrightarrow{\boldsymbol{J}} \) describes how electric current is distributed within a conductor. It's a vector quantity, revealing not just how much current exists, but also its direction.
In this case, the current density within the cylinder varies with radius \( r \) according to:\[ \overrightarrow{\boldsymbol{J}} = \frac{2 I_{0}}{\pi a^{2}} \left[1-\left(\frac{r}{a}\right)^{2}\right] \hat{k} \]This function shows that the current density decreases as you move from the center \( r=0 \) towards the outer edge \( r=a \) of the cylinder.

• At the cylinder’s center, the current density is at its maximum.
• At the surface \( r = a \), the current density drops to zero.

Understanding this variation is crucial for integrating appropriately later on.

Integral Calculus

Integral calculus is essential in determining the cumulative quantities from variable distributions. In this exercise, it's used to derive the current \( I \) passing through portions of the cylinder and to find \( I_0 \).
The integral for total current \( I_0 \) involves integrating the current density \( \overrightarrow{\boldsymbol{J}} \) over the cylinder's entire cross-sectional area:\[ \int_{0}^{a} \int_{0}^{2\pi} \overrightarrow{\boldsymbol{J}} \cdot r \, d\theta \, dr \]This requires finding the product of \( J \) with the infinitesimal area \( r \, d\theta \), integrating with respect to \( \theta \) and \( r \), which results in confirming that \( I_0 \) is indeed the total current.
Similarly, to find current for \( r \leq a \), a similar integral setup is needed, but it only covers up to radius \( r \), showing how much current contributes at lesser radii.

Cylinder Symmetry

Cylinder symmetry simplifies the math involved in problems like this one by acknowledging the geometry and symmetrical properties of the object. These properties make certain assumptions about current and field distributions reasonable.
Due to the azimuthal symmetry:

• The magnetic field \( \vec{B} \) at any point depends only on the radial distance \( r \) from the axis, simplifying calculations.
• The current density \( \overrightarrow{\boldsymbol{J}} \) is uniform along any cylindrical shell of radius \( r \), changing only with \( r \), not \( \theta \).

This symmetry is pivotal when using Ampere’s Law, as it allows the integral path to be a simple circle, making the integration of the magnetic field straightforward.
By recognizing and using cylinder symmetry, the mathematical process becomes more manageable and logical, reflecting physical reality effectively.