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 and varies according to the relationship $$ \begin{aligned} \overrightarrow{\boldsymbol{J}} &=\left(\frac{b}{r}\right) e^{(r-a) / \delta} \hat{\boldsymbol{k}} & \text { for } r \leq a \\ &=\mathbf{0} & \text { for } r \geq a \end{aligned} $$ where the radius of the cylinder is \(a=5.00 \mathrm{~cm}, r\) is the radial distance from the cylinder axis, \(b\) is a constant equal to \(600 \mathrm{~A} / \mathrm{m},\) and \(\delta\) is a constant equal to \(2.50 \mathrm{~cm} .\) (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 \(I_{0}\) (b) Using Ampere's law, derive an expression for the magnetic field \(\overrightarrow{\boldsymbol{B}}\) in the region \(r \geq a\). 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 \(\overrightarrow{\boldsymbol{B}}\) in the region \(r \leq \underline{a}\). (e) Evaluate the magnitude of the magnetic field at \(r=\delta, r=a,\) and \(r=2 a\)

Step by step solution

01

Calculate the Total Current

The total current \(I_{0}\) can be calculated by integrating the given current density \(\overrightarrow{\boldsymbol{J}}\) over the cross sectional area of the cylinder. This can be written as:\[I_{0} = \int_{0}^{a} 2\pi r J \,dr = \int_{0}^{a} 2 \pi r \left(\frac{b}{r}\right) e^{(r-a) / \delta} \,dr = 2\pi b \int_{0}^{a} e^{(r-a) / \delta} \,dr = 2\pi b \delta (e^{-a/\delta} - 1)\]

02

Apply Ampere's law to find \(\overrightarrow{\boldsymbol{B}}\) for \(r \geq a\)

Ampere's law, in integral form, states that the integral of the magnetic field \(\overrightarrow{\boldsymbol{B}}\) around a closed loop is equal to \(\mu_{0}\) times the total current enclosed by the loop. Therefore, for \(r \geq a\), since we only care about the current enclosed by the cylinder (which is \(I_{0}\)):\[\overrightarrow{\boldsymbol{B}} = \frac{\mu_{0} I_{0}}{2 \pi r} \hat{\boldsymbol{k}}\]

03

Calculate the Current within Radius \(r \leq a\)

Next, we find the current \(I\) enclosed by a cross section of radius \(r \leq a\). Similar to step 1, the current density is integrated over the cross sectional area of the loop:\[I = \int_{0}^{r} 2\pi r' J \, dr' = \int_{0}^{r} 2 \pi r' \left(\frac{b}{r'}\right) e^{(r'-a) / \delta} \,dr' = 2\pi b \delta (1 - e^{-r/\delta})\]

04

Apply Ampere's law to find \(\overrightarrow{\boldsymbol{B}}\) for \(r \leq a\)

Again by Ampere's law, for \(r \leq a\), we get:\[\overrightarrow{\boldsymbol{B}} = \frac{\mu_{0} I}{2 \pi r} \hat{\boldsymbol{k}}\]where \(I\) is the current enclosed by the loop radius \(r\).

05

Calculate the Magnitude of Magnetic Field

Substitute the respective value of \(r\) into the formulas obtained in step 2 and step 4 to calculate the magnetic field at \(r=\delta, r=a, r=2a\).

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Key Concepts

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

Current Density

Current density is a measure of how much electric current flows through a given area. It expresses the distribution of electric current across the cross-section of a conductor. In the case of the solid cylinder, the current density is given by the function \[\overrightarrow{\boldsymbol{J}} = \left(\frac{b}{r}\right) e^{(r-a) / \delta} \hat{\boldsymbol{k}} \quad \text{for } r \leq a\] - **Symmetry**: The current density is symmetric about the axis of the cylinder, meaning it behaves the same in all directions around the axis. This symmetry simplifies analysis and calculations.- **Variation**: Unlike some problems where current density is constant, here it decreases as you move away from the center towards the edge of the cylinder, due to the term \( e^{(r-a) / \delta}\). - **Integration for Total Current**: To find the total current, integrate the current density over the cross-sectional area:\[I_0 = \int_0^a 2\pi r \left(\frac{b}{r}\right) e^{(r-a) / \delta} \,dr\]This yields the expression \( I_0 = 2\pi b \delta (e^{-a/\delta} - 1)\), giving us the total current flowing through the cross-section of the cylinder.

Magnetic Field

The magnetic field arises from the movement of electric charges, such as the current flowing through our cylinder. Ampere's Law is a powerful tool used to derive expressions for magnetic fields in regions either inside or outside of current-carrying conductors.- **Outside Magnetic Field**: For points outside the cylinder (\(r \geq a\)), Ampere's Law relates the loop integral of the magnetic field around a closed path to the current enclosed by the path. By choosing a circular path around the cylinder and conducting the integration, the magnetic field is\[\overrightarrow{\boldsymbol{B}} = \frac{\mu_0 I_0}{2 \pi r} \hat{\boldsymbol{k}}\] This result shows the magnetic field decreases with distance from the cylinder.- **Inside Magnetic Field**: For inside the cylinder (\(r \leq a\)), the magnetic field expression derived shows\[\overrightarrow{\boldsymbol{B}} = \frac{\mu_0 I}{2 \pi r} \hat{\boldsymbol{k}}\],where \(I\) is part of the total current enclosed by the circle of radius \(r\).Understanding this helps us see how magnetic fields vary based on spatial positioning around the wire.

Cylinder Current

The concept of cylinder current is central to this problem. This involves considering the total current flowing through sections of the cylinder and the resulting magnetic field.- **Total vs. Enclosed Current**: - **Total Current** \(I_0\): Derived from integrating current density across the entire cross-section.- **Current Within Radius** (\(r \leq a\)) is calculated by integrating the current density over a cross-sectional area up to any radius \(r\):\[I = \int_0^r 2\pi r' \left(\frac{b}{r'}\right) e^{(r'-a) / \delta} \,dr'\] This results in\(I = 2\pi b \delta (1 - e^{-r/\delta})\).Knowing the fraction of the total current helps define the magnetic field magnitude. If this current \(I\) differs from \(I_0\), adjustments in field magnitude at radial positions need consideration. This distinction is paramount when applying Ampere's Law to regions within or outside the cylinder.