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 symmetrical about the cylinder axis, is not constant and varies according to the relationship $$ \begin{array}{rlrl}{\overrightarrow{\boldsymbol{J}}} & {=\left(\frac{b}{r}\right) e^{(r-a) / \delta} \hat{\boldsymbol{k}}} & {} & {\text { for } r \leq a} \\ {} & {=\mathbf{0}} & {} & {\text { for } \boldsymbol{r} \geq a}\end{array} $$ 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 \(\vec{B}\) 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 \(\vec{B}\) in the region \(r \leq a\) (e) Evaluate the magnitude of the magnetic field at \(r=\delta, r=a,\) and \(r=2 a\) .

Step by step solution

01

Expression for Total Current Iâ‚€

To find the total current \(I_{0}\), integrate the current density \(\vec{J}\) over the cross-sectional area of the cylinder. The area of a differential ring at radius \(r\) and thickness \(dr\) is \(2\pi r \, dr\). Thus, the differential current \(dI\) is given by:\[ dI = J(r) \cdot 2\pi r \, dr = \left(\frac{b}{r}\right) e^{(r-a)/\delta} \cdot 2\pi r \, dr \]\[ dI = 2\pi b e^{(r-a)/\delta} \, dr \]Integrate this expression from \(r = 0\) to \(r = a\) to get \(I_0\):\[ I_{0} = \int_{0}^{a} 2\pi b e^{(r-a)/\delta} \, dr \]To solve, set \(u = \frac{r-a}{\delta} \Rightarrow du = \frac{1}{\delta} \, dr \Rightarrow dr = \delta \, du\), then change limits accordingly. \(u= -\frac{a}{\delta}\to 0\).\[ I_{0} = 2\pi b \delta \int_{-\frac{a}{\delta}}^{0} e^{u} \, du = 2\pi b \delta [e^{u}]_{-\frac{a}{\delta}}^{0} \]\[ I_{0} = 2\pi b \delta (1 - e^{-a/\delta}) \]

02

Numerical Calculation of Iâ‚€

Substitute the given values into the expression for \(I_{0}\):- \(b = 600 \, \text{A/m}\)- \(\delta = 2.50 \, \text{cm} = 0.0250 \, \text{m}\)- \(a = 5.00 \, \text{cm} = 0.0500 \, \text{m}\)\[ I_{0} = 2\pi \times 600 \times 0.0250 (1 - e^{-\frac{0.0500}{0.0250}}) \]\[ I_{0} \approx 2\pi \times 600 \times 0.0250 \times (1 - e^{-2}) \]\[ I_{0} \approx 2\pi \times 600 \times 0.0250 \times (1 - 0.1353) \]\[ I_{0} \approx 2\pi \times 600 \times 0.0250 \times 0.8647 \]\[ I_{0} \approx 81.59 \, \text{A} \]

03

Magnetic Field Expression for r ≥ a

Use Ampere's Law for \(r \geq a\), which states \(\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed}\). Since \(r \ge a\), all the current \(I_0\) is enclosed:\[ B(2\pi r) = \mu_0 I_{0}\]\[ B = \frac{\mu_0 I_{0}}{2\pi r}\]This represents the magnetic field outside the cylinder in terms of \(I_{0}\).

04

Current Expression for r ≤ a

Find the current \(I\) in a cross-section of radius \(r\). Integrate the differential current \(dI\) from \(0\) to \(r\):\[ I = \int_{0}^{r} 2\pi b e^{(r'-a)/\delta} \, dr' \]Using substitution \(u = \frac{r'-a}{\delta}\), \(dr' = \delta \, du\), limits become \(u=-\frac{a}{\delta}\) to \(\frac{r-a}{\delta}\):\[ I = 2\pi b \delta [e^{u}]_{-\frac{a}{\delta}}^{\frac{r-a}{\delta}} \]\[ I = 2\pi b \delta (e^{(r-a)/\delta} - e^{-a/\delta}) \]Since \(I_0 = 2\pi b \delta (1 - e^{-a/\delta})\), express \(I\) in terms of \(I_0\):\[ I = I_0 \frac{e^{(r-a)/\delta} - e^{-a/\delta}}{1 - e^{-a/\delta}} \]

05

Magnetic Field Expression for r ≤ a

Apply Ampere's Law, \(\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed} = \mu_0 I\).\[ B(2\pi r) = \mu_0 I_0 \frac{e^{(r-a)/\delta} - e^{-a/\delta}}{1 - e^{-a/\delta}} \]\[ B = \frac{\mu_0 I_0}{2\pi r} \frac{e^{(r-a)/\delta} - e^{-a/\delta}}{1 - e^{-a/\delta}} \]This is the expression for the magnetic field \(\vec{B}\) for \(r \le a\).

06

Evaluation of Magnetic Field at Specific Points

Calculate \(B\) at \(r = \delta, r = a, \) and \(r = 2a\):\(r = a:\)\[ B = \frac{\mu_0 I_0}{2\pi a} \]\(r = 2a:\)\[ B = \frac{\mu_0 I_0}{4\pi a} \] \(r = \delta:\)Use \(r \le a\) formula:\[ B = \frac{\mu_0 I_0}{2\pi \delta} \frac{e^{0} - e^{-a/\delta}}{1 - e^{-a/\delta}} = \frac{\mu_0 I_0}{2\pi \delta} \]These calculations give the magnitudes of the magnetic field at different distances \(r\).

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Key Concepts

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

Magnetic Field in a Cylinder

The concept of a magnetic field inside a cylindrical conductor involves the study of how the magnetic field behaves both inside and outside the cylinder. In our specific situation, we consider a long solid cylinder carrying an electrical current over its cross-sectional area. The current generates a magnetic field due to the movement of charges within the conductor.
As we look at the cylinder, the magnetic field outside the cylinder \(r \geq a\) can be determined using Ampere's Law. Ampere's Law is a powerful tool used to relate the magnetic field along a closed loop to the current flowing through that loop, expressed mathematically as \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed}\.- For \(r \geq a\), all the current in the cylinder contributes to creating the magnetic field. Thus, the magnetic field \B\ is cylindrically symmetric, and Ampere's Law simplifies to \B = \frac{\mu_0 I_0}{2\pi r}\, ensuring calculation simplicity.- Inside the cylinder \(r \leq a\), the magnetic field can vary depending on how the current is distributed within the conductor. Since only part of the total current is enclosed within radius \r\, we adjust Ampere's Law to account for this smaller area.

Current Density Distribution

Understanding current density distribution within a cylinder is key to calculating related magnetic fields. Current density, denoted as \(\vec{J}\), describes how much current flows through a unit area of the conductor. In this exercise, the current density \(\vec{J}\) is not uniform but varies with radial distance \(r\) from the axis, and described by the equation:\(\vec{J} = \left(\frac{b}{r}\right) e^{(r-a)/\delta} \hat{k}\)- As \(r\) increases from the center to the edge \(a\), \(\vec{J}\) changes in a non-linear fashion due to the \(e^{(r-a)/\delta}\) factor.
- \(b\) and \(\delta\) are constants that adjust how steeply \(\vec{J}\) drops off; this suggests that near the center, the current density is larger, decreasing with radial distance.- Beyond the cylinder's radius \(a\), there is no current because \(\vec{J} = 0\) for \(r \geq a\).This variable distribution affects how our calculations for both total and partial currents are approached, relying on integration methods over the radial extent.

Integration of Vector Fields

Calculating current and related properties in the presence of variable distributions often necessitates integrating vector fields. Here, the task is to integrate the non-uniform current density \(\vec{J}\) over the cross-sectional area to determine the total current \(I_0\).- The differential current, \(dI\), is derived from \(\vec{J}\) and the differential area element \(2\pi r \, dr\), giving \(dI = \left(\frac{b}{r}\right) e^{(r-a)/\delta} \cdot 2\pi r \, dr = 2\pi b e^{(r-a)/\delta} \, dr\).- This differential current is integrated from \(r = 0\) to \(r = a\), using a substitution for simplification, resulting in the expression for the total current \(I_0 = 2\pi b \delta (1 - e^{-a/\delta})\).This integration approach is critical for both determining the total current through the cylinder and understanding how much current is within a specific radius, directly influencing the magnetic field distribution.