Problem 75 - 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 \(\overrightarrow{J}\). The current density, although symmetric about the cylinder axis, is not constant and varies according to the relationship $$\overrightarrow{J} = (\frac{b}{r})e^{(r a)/\delta}\hat{k} \space for \space r \leq a$$ $$=0 \space for \space r \geq a$$ where the radius of the cylinder is a = 5.00 cm, \(r\) is the radial distance from the cylinder axis, \(b\) is a constant equal to 600 A/m, and \(\delta\) is a constant equal to 2.50 cm. (a) Let \(I_0\) be the total current passing through the entire cross section of the wire. Obtain an expression for \(I_0\) in terms of \(b\), \(\delta\), and a. Evaluate your expression to obtain a numerical value for I0. (b) Using Ampere's law, derive an expression for the magnetic field \(\overrightarrow{B}\) in the region \(r \leq a\). Express your answer in terms of \(I_0\) rather than b. (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. Express your answer in terms of \(I_0\) rather than b. (d) Using Ampere's law, derive an expression for the magnetic field \(\overrightarrow{B}\) in the region \(r \leq a\). (e) Evaluate the magnitude of the magnetic field at \(r = \delta\), \(r = a\), and \(r = 2a\).

Step-by-Step Solution

Answer

(a) Derive I0 using integral of J; (b) B = (μ0 I0)/(2πr); (c) I_r in terms of I0; (d) Express B using I_r; (e) Evaluate B at given r.

Step 1: Calculate the Total Current I_0 Through the Cylinder

To find the total current \( I_0 \), we integrate the current density \( \overrightarrow{J} \) over the cross-sectional area of the cylinder. The current density is given by \( \overrightarrow{J} = \left(\frac{b}{r}\right)e^{\frac{ra}{\delta}}\hat{k} \) for \( r \leq a \). Therefore, we calculate:\[ I_0 = \int_0^{2\pi} \int_0^a J(r) \cdot r \, dr \, d\theta = \int_0^{2\pi} \int_0^a \left(\frac{b}{r}\right)e^{\frac{ra}{\delta}} \cdot r \, dr \, d\theta \]The expression simplifies to:\[ I_0 = 2\pi b \int_0^a e^{\frac{ra}{\delta}} \, dr \]Solving the integral:\[ I_0 = \frac{2\pi b \delta}{a}(e^{\frac{a^2}{\delta}} - 1) \] Substitute values: \( b = 600 \) A/m, \( a = 0.05 \) m, \( \delta = 0.025 \) m:\[ I_0 = \frac{2\pi \times 600 \times 0.025}{0.05}(e^{\left(\frac{0.05^2}{0.025}\right)} - 1) \]Evaluate further to find the numerical value for \( I_0 \).

Step 2: Use Ampere's Law to Derive Magnetic Field B for r ≤ a

Using Ampere's Law in the region \( r \leq a \), the magnetic field \( B \) is calculated using:\[ \oint \overrightarrow{B} \cdot d\overrightarrow{l} = \mu_0 I_{ ext{inside}} \]For a circular path of radius \( r \) centered on the axis:\[ B(2\pi r) = \mu_0 I \]Thus:\[ B = \frac{\mu_0 I}{2\pi r} \]Express \( I \) as a fraction of \( I_0 \) using \( I_{r} = I_0 \frac{\int_0^r (b/r') e^{(r'a)/\delta} \cdot r \, dr'}{\int_0^a (b/r') e^{(r'a)/\delta} \, dr'} \).

Step 3: Calculate Current I for r ≤ a

The current \( I_r \) inside a radius \( r \leq a \) is given by the integral of current density up to \( r \):\[ I_r = \int_0^{2\pi} \int_0^r J(r') \, r' \, dr' \, d\theta = 2\pi b \int_0^r e^{\frac{r'a}{\delta}} \, dr' \]Substitute and integrate:\[ I_r = \frac{2\pi b \delta}{a}(e^{\frac{r^2}{\delta}} - 1) \] Relate \( I_r \) to \( I_0 \) using: \[ I_r = I_0 \frac{e^{\frac{r^2}{\delta}} - 1}{e^{\frac{a^2}{\delta}} - 1} \]

Step 4: Determine B in Terms of I_0 for r ≤ a

From Ampere's Law, substitute the expression for \( I_r \):\[ B = \frac{\mu_0 I_r}{2\pi r} \]Substitute the expression for \( I_r \) in terms of \( I_0 \):\[ B = \frac{\mu_0 I_0}{2\pi r} \frac{e^{\frac{r^2}{\delta}} - 1}{e^{\frac{a^2}{\delta}} - 1} \]

Step 5: Evaluate Magnitude of B at Specific Values of r

Calculate \( B \) at specified \( r \) values using the derived expression:- For \( r = \delta \): \[ B_{\delta} = \frac{\mu_0 I_0}{2\pi \delta} \frac{e^{\frac{\delta^2}{\delta}} - 1}{e^{\frac{a^2}{\delta}} - 1} \] - For \( r = a \): \[ B_{a} = \frac{\mu_0 I_0}{2\pi a} \]- For \( r = 2a \) (outside the cylinder where B would conventionally be 0): Consider the current passed through the entire cylinder.

Key Concepts

Current Density

Current density is a measure of the electric current flowing through a unit area of a material. It is a vector quantity, representing the direction and magnitude of the current at a certain point. In the context of our exercise, the current density within the cylinder is given by \( \overrightarrow{J} = \left(\frac{b}{r}\right)e^{(\frac{ra}{\delta})}\hat{k} \) for \( r \leq a \). Here, the variable \( r \) represents the radial distance from the center of the cylinder.

• **Formula Components:**
  • \( b \): A constant representing current capacity, given as 600 A/m.
  • \( e^{(\frac{ra}{\delta})} \): An exponential factor influencing how the current density changes with distance \( r \).
  • \( \hat{k} \): Denotes direction along the cylinder's axis, emphasizing current flow along the cylinder's length.
• **Behavior:** The current density decreases radially from the interior to the surface of the cylinder, dictated by \( \frac{b}{r} \) and exponentially by \( e^{(\frac{ra}{\delta})} \).

It is crucial to understand that current density significantly affects how we calculate the overall current passing through any cross-section of the wire, allowing us to analyze the magnetic field using Ampere's Law.

Magnetic Field

A magnetic field refers to a region around a moving electric charge where magnetic forces can be detected. When current flows through a conductor like our cylinder, it creates a magnetic field around it. We use Ampere's Law to calculate this field:Ampere's Law states that the line integral of the magnetic field \( \overrightarrow{B} \) around any closed path is equal to \( \mu_0 \) times the current enclosed by that path:\[ \oint \overrightarrow{B} \cdot d\overrightarrow{l} = \mu_0 I_\text{enclosed} \]For - a circle centered on the cylinder with radius \( r \leq a \):- The magnetic field lines are circular and concentric.- By symmetry: \( B(2\pi r) = \mu_0 I_r \) leading to \( B = \frac{\mu_0 I_r}{2\pi r} \).- When expressed in terms of total current \( I_0 \), \( I_r \) is calculated to find \( B \), considering the entire flow in sectors up to \( r \).Understanding this helps us evaluate the effects of varying \( r \) on the magnetic field's strength and direction, which can differ at points like \( r = \delta \) or on the conductor's boundary \( r = a \).

Integrals in Cylindrical Coordinates

Integrating in cylindrical coordinates is vital for computing areas and volumes related to circular or cylindrical objects. This technique is particularly useful for problems involving symmetries about a central axis, such as the one demonstrated in our exercise.### Integral SetupCylindrical coordinates

• use the radius \( r \), angular position \( \theta \), and height \( z \) for integration, replacing traditional Cartesian methods.
• The expressions consider the circular cross-section nature, hence use \( dr \), \( d\theta \), and \( dz \) for integrals.

### Specific ApplicationThe exercise involves integrating the current density over a cylindrical shape:- The radial integral: \( \int_0^a J(r) \, r \, dr \) computes current contribution from core to surface.- Another angular component: \( \int_0^{2\pi} d\theta \), capturing circular symmetry around the axis.The conversion simplifies to handle exponential variability with radial factors directly, demonstrating how radial dependencies, like \( \frac{b}{r} \, \times e^{(ra/\delta)} \) affect integration results.By mastering these integral techniques, you gain the tools necessary to evaluate complex magnetic and electric fields in advanced physics and engineering scenarios.