Problem 77 - University Physics with Modern Physics

Question

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

Answer

The results show continuity of the magnetic field at the boundary: \( B = \frac{\mu_0 I_0}{2\pi a} \) for both \( r=a \) cases.

Step 1: Integrate Current Density Over Cylindrical Cross-Section

To find the total current \( I_0 \), integrate the current density \( \vec{J} \) over the entire cross-sectional area of the cylinder where \( r \leq a \). The current passing through a differential area element \( dA = 2\pi r \, dr \) is given by \( dI = J \, dA \). Hence,\[ I_0 = \int_0^a \left( \frac{2I_0}{\pi a^2} \right) \left( 1 - \frac{r^2}{a^2} \right) \cdot 2\pi r \, dr \].

Step 2: Evaluate the Integral

Evaluating the integral calculated in Step 1:\[I_0 = \frac{4I_0}{a^2} \int_0^a \left( r - \frac{r^3}{a^2} \right) dr = \frac{4I_0}{a^2} \left[ \frac{r^2}{2} - \frac{r^4}{4a^2} \right]_0^a = \frac{4I_0}{a^2} \left( \frac{a^2}{2} - \frac{a^4}{4a^2} \right)\]This simplifies to:\[ I_0 = 2I_0 \left( \frac{1}{2} - \frac{1}{4} \right) = I_0 \].

Step 3: Apply Ampere's Law for Region r ≥ a

For \( r \geq a \), no current is enclosed in the Amperian loop, but Ampere's law tells us: \( \oint \vec{B} \cdot d\vec{l} = \mu_0 I_\text{enclosed} \). Since \( I_\text{enclosed} = I_0 \) for \( r \geq a \), the magnetic field \( \vec{B} \) is uniform on the path of integration:\[ B(2\pi r) = \mu_0 I_0 \Rightarrow B = \frac{\mu_0 I_0}{2\pi r} \].

Step 4: Calculate Enclosed Current for r ≤ a

To calculate the current \( I(r) \) contained in a circular area of radius \( r \leq a \), integrate \( J \) over \( r \leq a \):\[ I(r) = \int_0^r \left( \frac{2I_0}{\pi a^2} \right) \left( 1 - \frac{r'^2}{a^2} \right) \, 2\pi r' \, dr' \].

Step 5: Evaluate the Integral to Find Enclosed Current

Evaluate the integral derived in Step 4:\[ I(r) = \frac{4I_0}{a^2} \left[ \frac{r'^2}{2} - \frac{r'^4}{4a^2} \right]_0^r = \frac{4I_0}{a^2} \left( \frac{r^2}{2} - \frac{r^4}{4a^2} \right) \].

Step 6: Apply Ampere's Law for Region r ≤ a

For \( r \leq a \), Ampere's law gives \( \oint \vec{B} \cdot d\vec{l} = \mu_0 I(r) \):\[ B(2\pi r) = \mu_0 \cdot \frac{4I_0}{a^2} \left( \frac{r^2}{2} - \frac{r^4}{4a^2} \right)\]. Thus,\[ B = \frac{\mu_0 I_0 r}{2\pi a^2} \left( 1 - \frac{r^2}{a^2} \right) \].

Step 7: Compare Magnetic Fields at r = a

Both expressions for \( \vec{B} \) when evaluated at \( r = a \) are equal:For \( r \geq a \), \( B = \frac{\mu_0 I_0}{2\pi a} \).For \( r \leq a \),\( B = \frac{\mu_0 I_0}{2\pi a} \).Both methods give the same result at \( r = a \), confirming the continuity of the magnetic field at this boundary.

Key Concepts

Understanding Current Density

Current density is an important concept when analyzing electrical currents, especially in complex systems like a cylindrical conduit. It's a measure of the amount of electric current flowing per unit area of cross-section through a material. In our cylindrical example, the current density \( \vec{J} \) is not uniform but varies with radial distance \( r \). The formula given is \( \vec{J} = \frac{2 I_{0}}{\pi a^{2}} \left[1-\left(\frac{r}{a}\right)^{2}\right] \hat{k}\), where \( a \) is the cylinder's radius, and \( I_0 \) is a constant. This equation tells us that the current density decreases with the square of the distance from the center of the cylinder, meaning that closer to the axis, more current is flowing.

• The function's peak at \( r = 0 \) indicates maximum current density at the center.
• As \( r \) approaches \( a \), the current density becomes zero, conforming to the physical boundary of the cylinder.

Deeper understanding of current density helps in determining the total current flowing through the wire by integrating across the cross-sectional area.

Exploring the Magnetic Field

The magnetic field \( \vec{B} \) is a vector field surrounding electric currents and is described by Ampere's Law. For our cylindrical conductor, the current flowing induces a magnetic field. To determine this field in the region \( r \geq a \), we apply Ampere's Law: \ \( \oint \vec{B} \cdot d\vec{l} = \mu_0 I_\text{enclosed} \), where \( I_\text{enclosed} \) is the total current enclosed by the loop. This approach reveals that the magnetic field decreases with distance as \ \( B = \frac{\mu_0 I_0}{2\pi r} \). \

• In the region outside the cylinder, the entire current \( I_0 \) contributes to the magnetic field.
• The field strength inversely depends on \( r \), signifying it weakens further from the current source.

Understanding the magnetic field provides insights into how electrical currents influence space around them, which is pivotal in the design of electrical devices.

Applying Cylindrical Coordinates

Cylindrical coordinates are a system of geometry that extends polar coordinates by adding a height dimension for three-dimensional space. This is particularly useful for problems with symmetrical cylindrical shapes where the geometry naturally aligns with the system. The parameters include the radial distance \( r \), angle \( \theta \), and height \( z \).

• This coordinate system simplifies the integration process within cylindrical geometries.
• For the given cylinder, \( r \) is the distance from the center, \( \theta \) is ignored due to symmetry, and \( z \) aligns with the cylinder's axis.

Embracing cylindrical coordinates enables clearer calculations and understanding of factors like current density distribution across the cylinder, thus aiding in solutions to electromagnetic problems.

Integration of Current Densities

Integrating current densities involves accumulating the contributions of current density over the cross-section to find the total current. For our problem, integration is executed over a circle's area for \ \( r \leq a \). The formula used is \ \( I_0 = \int_0^a \left( \frac{2I_0}{\pi a^2} \right) \left( 1 - \frac{r^2}{a^2} \right) \cdot 2\pi r \, dr \),

• The integral uses a polar area element \ \( dA = 2\pi r \, dr \, \ \).
• This accumulation across all differential elements gives the total current flow through the conductor.

By solving this integral correctly, we confirm the supplied total current \( I_0 \), ensuring that the calculations conform to physical reality and validating the distribution's mathematical model.