Problem 71 - University Physics with Modern Physics

Question

A long, straight wire with a circular cross section of radius \(R\) carries a current \(L\) . Assume that the current density is not constant across the cross section of the wire, but rather varies as \(J=\alpha r,\) where \(\alpha\) is a constant. (a) By the requirement that \(J\) integrated over the cross section of the wire gives the total current \(I,\) calculate the constant \(\alpha\) in terms of \(I\) and \(R\) (b) Use Ampere's law to calculate the magnetic field \(B(r)\) for (i) \(r \leq R\) and (ii) \(r \geq R\) . Express your answers in terms of \(I\) .

Step-by-Step Solution

Answer

\(\alpha = \frac{3I}{2\pi R^3}; B(r) = \frac{\mu_0 I r^2}{2\pi R^3}, r \leq R; B(r) = \frac{\mu_0 I}{2\pi r}, r \geq R.\)

Step 1: Find Expression for Total Current

We start with the current density defined as \( J = \alpha r \). We need to integrate this over the circular cross-section to find the total current \( I \). The differential area element in cylindrical coordinates is \( dA = 2\pi r \, dr \). Therefore, the total current \( I \) is given by:\[I = \int_0^R J \, dA = \int_0^R \alpha r (2\pi r \, dr)\]Simplifying gives:\[I = 2\pi \alpha \int_0^R r^2 \, dr\]

Step 2: Solve the Integral

Now, we need to evaluate the integral:\[\int_0^R r^2 \, dr = \left[ \frac{r^3}{3} \right]_0^R = \frac{R^3}{3}\]So, substitute this result back into the expression for \( I \):\[I = 2\pi \alpha \left( \frac{R^3}{3} \right) = \frac{2\pi \alpha R^3}{3}\]

Step 3: Solve for Alpha

To find the constant \( \alpha \), rearrange the equation:\[\alpha = \frac{3I}{2\pi R^3}\]

Step 4: Magnetic Field Inside the Wire: Ampere's Law

For \( r \leq R \), use Ampere's Law, \( \oint B \, dl = \mu_0 I_{ ext{enc}} \). The path of integration is a circle of radius \( r \). \( B \) is constant along this path, so the left side becomes \( B(2\pi r) \).The enclosed current \( I_{\text{enc}} \) is the integral of \( J \) from \( 0 \) to \( r \):\[I_{\text{enc}} = \int_0^r \alpha r' (2\pi r') \, dr' = 2\pi \alpha \int_0^r r'^2 \, dr'\]Evaluate the integral:\[\int_0^r r'^2 \, dr' = \frac{r^3}{3}\]Substitute back:\[I_{\text{enc}} = \frac{2\pi \alpha r^3}{3}\]

Step 5: Solve for Magnetic Field Inside the Wire

Using Ampere's Law:\[B(2\pi r) = \mu_0 \frac{2\pi \alpha r^3}{3}\]Solve for \( B \):\[B = \frac{\mu_0 \alpha r^2}{3}\]Substituting \( \alpha \) from Step 3:\[B = \frac{\mu_0 (3I) r^2}{3(2\pi R^3)} = \frac{\mu_0 I r^2}{2\pi R^3}\]

Step 6: Magnetic Field Outside the Wire: Ampere's Law

For \( r \geq R \), the enclosed current \( I_{\text{enc}} \) is the total current \( I \). Using Ampere's Law again:\[B(2\pi r) = \mu_0 I\]Solve for \( B \):\[B = \frac{\mu_0 I}{2\pi r}\]

Key Concepts

Current Density

In the study of electromagnetism, the current density, denoted by \( J \), is an important concept. Current density refers to the flow of electric charge per unit area through a cross section. In the given exercise, the current density is not constant and is expressed as \( J = \alpha r \). Here, \( \alpha \) is a constant that represents the relationship between the current density and the distance \( r \) from the center of the wire's cross section. This implies that the current density increases linearly with distance from the center.Understanding current density is crucial when dealing with varying conductors as it allows us to determine how current is distributed within a conductor. Calculating \( \alpha \) requires integrating the given current density over the entire cross section and equating it to the total current \( I \). This integral helps clarify how the density varies and impacts the total current flowing through the wire.

Magnetic Field

Magnetic fields arise from the movement of electric charges, and understanding their behavior is critical in radiating magnetic phenomena. Ampere's Law, which relates magnetic fields to the current that produces them, is particularly useful here. The magnetic field \( B \), surrounding a current-carrying wire, varies with the radial distance \( r \) from the wire.For points

• inside the wire \( (r \leq R) \), Ampere's Law is applied using a path of integration, a circle with radius \( r \) inside the wire.
• outside the wire \( (r \geq R) \), the magnetic field depends solely on the total current \( I \), not on how it is distributed inside the wire.

Using these conditions, the exercise calculates the magnetic field inside (\( B = \frac{\mu_0 I r^2}{2\pi R^3} \)) and outside (\( B = \frac{\mu_0 I}{2\pi r} \)) the wire, highlighting the inverse relationship with distance for points outside.

Cylindrical Coordinates

The use of cylindrical coordinates simplifies the integration processes required in electromagnetic problems involving symmetrical configurations like wires. In cylindrical coordinates:

• the distance from the axis is represented by \( r \),
• the angle around the axis by \( \theta \), and
• the height along the axis by \( z \).

This coordinate system is well-suited for calculations involving cylindrical objects, such as the long straight wire described in the exercise. Particularly, the differential area element used during integration is expressed as \( dA = 2\pi r \, dr \).Cylindrical coordinates streamline the calculation of integrals over circular cross-sections, allowing for simplification of the problem and ease in solving the differential equations that describe the behavior of electrical fields and currents in these geometries.

Integration in Physics

Integration is a fundamental mathematical tool used extensively in physics to solve problems involving continuous distributions of quantities, such as charge or current densities. In this exercise, integration is used to determine both the total current from the current density and the magnetic field using Ampere's Law.Highlights of integration in this context include:

• Calculating total current by integrating the current density \( J = \alpha r \) over the wire's cross-section.
• Using integration to find the enclosed current \( I_{\text{enc}} \) necessary for determining the magnetic field inside specific regions.

These integrations transform physical quantities that vary over an area or volume into useful scalar quantities like total current, facilitating deeper insights into electromagnetic phenomena. Understanding the integral's role in connecting mathematical descriptions to physical reality is key to effectively applying it in such physics problems.