Problem 71 - 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 but varies according to the relationship $$\overrightarrow{J} = \frac{2I_0}{\pi{a}^2} [1-(\frac{r}{a})^2]\hat{k} \space for \space r \leq a$$ $$=0 \space for \space r \geq a$$ 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 \(\overrightarrow{B}\) 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 \(\overrightarrow{B}\) 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 total current is \(I_0\). Use Ampère's law to find \(B\) outside and inside, ensuring continuity at \(r = a\). The results at \(r = a\) match.

Step 1: Understanding Current Density

The current density in the problem is given by the expression \( \overrightarrow{J} = \frac{2I_0}{\pi{a}^2} [1-(\frac{r}{a})^2]\hat{k} \). This density is radially symmetric and decreases non-linearly from the axis of the cylinder (\(r = 0\)) to its surface (\(r = a\)). No current flows outside of the cylinder (\(r \geq a\)).

Step 2: Total Current Through Cross Section

To find the total current, integrate the current density across the cylinder's cross-sectional area. \[I = \int_0^a \int_0^{2\pi} \overrightarrow{J} \cdot d\mathbf{A} = \int_0^a \int_0^{2\pi} \frac{2I_0}{\pi a^2} \left[1-\left(\frac{r}{a}\right)^2\right] r dr d\theta\]Calculating this, the integral over \(\theta\) gives a factor of \(2\pi\), and the \(r\) integral simplifies using substitution or direct integration to yield \(I = I_0\).

Step 3: Magnetic Field Outside Cylinder (r ≥ a)

For \(r \geq a\), use Ampère's Law: \[\oint \overrightarrow{B} \cdot d\overrightarrow{l} = \mu_0 I_{enc}\]For a circular path of radius \(r\), symmetry allows |\( \overrightarrow{B} \)| to be constant over the path length, giving:\[B(2\pi r) = \mu_0 I_0 \Rightarrow B = \frac{\mu_0 I_0}{2 \pi r}\]

Step 4: Current in Cross Section r ≤ a

For \(r \leq a\), calculate the enclosed current by integrating the current density over the cross-section of radius \(r\): \[I = \int_0^r \int_0^{2\pi} \frac{2I_0}{\pi a^2} \left[1-\left(\frac{r}{a}\right)^2\right] r dr d\theta\]This evaluates to a function of \(r\), specifically \(I = I_0 \left(\frac{r^2}{a^2} \right)\left(1 - \frac{r^2}{2a^2}\right)\).

Step 5: Magnetic Field Inside Cylinder (r ≤ a)

Using Ampère's Law inside the cylinder for \(r \leq a\), we use the same process: \[\oint \overrightarrow{B} \cdot d\overrightarrow{l} = \mu_0 I_{enc}\]Thus, \[B(2\pi r) = \mu_0 I_{enc}\]Solving with \(I_{enc}\) from the previous step gives a linearly increasing function of \(B(r)\), specific form requiring integration of the expression derived for \(I_{enc}\).

Step 6: Comparison at Boundary r = a

When \(r = a\), check that both expressions for \(B\) inside and outside yield the same value. The continuity of \(B\) is assured due to conservation laws. Verifying, they should show that \(B(a)\) is consistent across boundary regions.

Key Concepts

Understanding Current Density

Current density is a crucial concept in electromagnetism. It describes how electric current is distributed over an area. In this exercise, it is defined as \( \overrightarrow{J} = \frac{2I_0}{\pi{a}^2} [1-(\frac{r}{a})^2]\hat{k} \), where the symbol \( r \) represents the radial distance from the cylinder's center, and \( a \) is the radius of the cylinder. This expression tells us that the current density is not uniform across the cylinder; it's highest at the center and decreases non-linearly towards the edge of the cylinder. Understanding the behavior of current density helps in determining how much current passes through a particular section of the material.

Magnetic Field and Ampere's Law

The magnetic field surrounding a wire or a current-carrying cylinder is determined by Ampere's Law, which states that the line integral of the magnetic field \( \overrightarrow{B} \) around a closed path is equal to \( \mu_0 \) times the enclosed current (\( I_{enc} \)). For example, outside a cylinder (region \( r \geq a \)), we assume symmetry, which simplifies calculations. The expression \( B(2\pi r) = \mu_0 I_0 \) results in:

• \( B = \frac{\mu_0 I_0}{2 \pi r} \)

Inside the cylinder (region \( r \leq a \)), the field calculation involves determining how much current is enclosed within a circle of radius \( r \). Using Ampere's Law, we find:\[B(2\pi r) = \mu_0 I_{enc}\]This relationship helps derive the magnetic field inside depending on the radius.

Cylinder Symmetry

The concept of cylinder symmetry comes into play when analyzing physical situations involving cylinders. In electromagnetism, symmetry can significantly simplify problems—like in this exercise where the current density and resulting magnetic fields are considered symmetrical around the cylindrical axis. This makes calculations easier by allowing the use of symmetrical paths for integration, directly linking the radial and angular components of the system. Additionally, symmetry simplifies many physical laws, reducing a three-dimensional problem to a one-dimensional one, as only radial (\( r \)) variations need to be accounted for explicitly.

Integral Calculus in Electromagnetism

Integral calculus is fundamental in electromagnetism, especially when calculating quantities distributed over areas or volumes, like current and magnetic fields. For this exercise, integrals are used to find the total current by integrating the current density over the cylinder's cross-sectional area. The expression\[I = \int_0^a \int_0^{2\pi} \overrightarrow{J} \cdot d\mathbf{A}\]demands understanding of integrals over polar coordinates, where the angle \( \theta \) and radius \( r \) are pivotal. In this context, integrals help transition from local properties (current densities) to global properties (total current and magnetic fields). This continuous summation process is a powerful tool for connecting differential properties with total quantities in physical systems.