91Ó°ÊÓ

A long, straight, solid cylinder, oriented with its axis in the \(z\) -direction, carries a current whose current density is \(\vec{J}\) . The current density, although symmetric about the cylinder axis, is not constant but varies according to the relationship $$ \begin{aligned} \vec{J} &=\frac{2 I_{0}}{\pi a^{2}}\left[1-\left(\frac{r}{a}\right)^{2}\right] \hat{k} & \text { for } r \leq a \\ &=0 \quad \text { for } r \geq a \end{aligned} $$ 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

Find Total Current for r ≤ a

To find the total current \(I_0\) through the wire's cross-section, integrate the current density \( \vec{J} \) over the cross-sectional area of the cylinder. The total current \( I_0 \) is given by\[I_0 = \int_0^a \int_0^{2\pi} \vec{J} \cdot \hat{k} \, r \, d\theta \, dr = \int_0^a \int_0^{2\pi} \frac{2 I_0}{\pi a^2} \left[1 - \left(\frac{r}{a}\right)^2\right] \, r \, d\theta \, dr.\]Solving the integral, we first integrate over \(\theta\) from \(0\) to \(2\pi\):\[I_0 = \int_0^a \left( \frac{2 I_0}{\pi a^2} \left[1 - \left(\frac{r}{a}\right)^2\right] \, 2\pi r \right)\, dr.\]Integrate the resulting expression with respect to \(r\):\[I_0 = \frac{4 I_0}{a^2} \left( \int_0^a \left[r - \frac{r^3}{a^2}\right] \, dr \right) = \frac{4 I_0}{a^2} \left[\frac{r^2}{2} - \frac{r^4}{4a^2}\right]_0^a = I_0.\]Thus, the total current across the cross-section is indeed \(I_0\).

02

Apply Ampere's Law for r ≥ a

For \( r \geq a \), use Ampere's Law, \(\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc} \). As \( r \geq a \), the enclosed current \( I_{enc} = I_0 \). Ampere's loop is a circle of radius \( r \), so the path integral is\[B(2\pi r) = \mu_0 I_0.\]Thus, the magnetic field \( B \) is\[B = \frac{\mu_0 I_0}{2\pi r}\]for \( r \geq a \).

03

Find Current I for r \, leq a

For a circle of radius \( r \leq a \), integrate the current density \( \vec{J} \) from \(0\) to \(r\) to find the enclosed current \( I \):\[I = \int_0^r \int_0^{2\pi} \frac{2 I_0}{\pi a^2} \left[1 - \left(\frac{r'}{a}\right)^2\right] \, r' \, d\theta \, dr'.\]Evaluating\[I = \frac{4 I_0}{a^2} \int_0^r \left[r' - \frac{r'^3}{a^2}\right] \pi \, dr' = \frac{4 I_0}{a^2} \left[\frac{r'^2}{2} - \frac{r'^4}{4a^2}\right]_0^r = \frac{2I_0}{a^2} \left(r^2 - \frac{r^4}{2a^2}\right).\]

04

Apply Ampere's Law for r ≤ a

Using Ampere's Law for \( r \leq a \), for a circle of radius \( r \), the integral becomes\[B(2\pi r) = \mu_0 I.\]Using the expression for \( I \) found in Step 3:\[B(2\pi r) = \mu_0 \left(\frac{2I_0}{a^2}\right) \left(r^2 - \frac{r^4}{2a^2}\right).\]Solving for \( B \):\[B = \frac{\mu_0 I_0}{\pi a^2} \left(r - \frac{r^3}{2a^2}\right)\]for \( r \leq a \).

05

Compare Results at r = a

At \( r = a \), compare the magnetic fields derived in Steps 2 and 4:\[\text{For } r \geq a, \quad B = \frac{\mu_0 I_0}{2\pi a}.\]\[\text{For } r \leq a, \quad B = \frac{\mu_0 I_0}{\pi a} \left(1 - \frac{a^2}{2a^2}\right) = \frac{\mu_0 I_0}{2\pi a}.\]Both expressions yield the same magnetic field at \( r = a \).

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Key Concepts

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

current density

Current density (\(\vec{J}\)) tells us how much electric current is flowing through a unit area of a material. It's expressed as a vector, indicating not only the magnitude but also the direction of flow. In the case of the solid cylinder, the current density is symmetric about the axis but changes with distance from the center. This variance is described using the formula:

• \(\vec{J} = \frac{2 I_{0}}{\pi a^{2}}\left[1-\left(\frac{r}{a}\right)^{2}\right]\hat{k}\) for \(r \leq a\)
• \(\vec{J} = 0\) for \(r \geq a\)

This means that near the axis, the current density is higher, but as you move toward the outer surface of the cylinder, it decreases to zero. Understanding how current density distributes helps in calculating the total current flowing across the cylinder's cross-section.

magnetic field

The magnetic field (\(\vec{B}\)) is a vector field around a magnet or current-carrying conductor, representing magnetic influence. The magnetic field's direction is perpendicular to the current flow. Using Ampere's Law, one can calculate this field's strength and orientation. Ampere's Law states:\[\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}\]Where \(I_{enc}\) is the current enclosed by the loop. For our cylinder:

• For \(r \geq a\): The formula derived is \(B = \frac{\mu_0 I_0}{2\pi r}\).
• For \(r \leq a\): The magnetic field is expressed as \(B = \frac{\mu_0 I_0}{\pi a^2} \left(r - \frac{r^3}{2a^2}\right)\).

At \(r = a\), both expressions provide the same magnetic field strength, showing consistency in calculations and illustrating that the external field around the cylinder behaves uniformly.

cylinder cross-section

When dealing with a cylindrical object like this one, the cross-section is essentially a slice through the cylinder perpendicular to its axis. This circular cross-section is crucial for calculating current and magnetic fields. The radius \(a\) defines the boundary of the cylinder. The varying current density across this cross-section plays a significant role in determining the overall behavior of the magnetic field.Using the given current density, you can compute the current flowing through the cylinder's entire cross-section by integrating over the radial distance from the center to the edge (radius \(a\)). This approach uses the symmetry of the cylinder and helps to simplify calculations.

enclosed current

The enclosed current \(I_{enc}\) is the amount of current passing through a specific area within a conductor. It's important for applying Ampere's Law to find the magnetic field. For the given cylinder symmetry, the enclosed current varies with the distance \(r\) from the axis.To find the current contained within any circle of radius \(r\leq a\) at the cylinder's center, integrate the current density over the desired circular area. The formula becomes:\[I = \frac{2I_0}{a^2} \left(r^2 - \frac{r^4}{2a^2}\right)\]This integration reflects how much current flows through a smaller section of the cylinder, which directly affects the strength and characteristics of the magnetic field inside and outside the cylinder.