Problem 69 - University Physics with Modern Physics

Question

A long, straight wire with a circular cross section of radius \(R\) carries a current \(I\). Assume that the current density is not constant across the cross section of the wire, but rather varies as \(J =\) \(ar\), where a 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 \(a\) 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

(a) \(a = \frac{3I}{2\pi R^3}\); (b) For \(r \leq R\), \(B = \frac{3Ir^2}{4\pi R^3}\); for \(r \geq R\), \(B = \frac{I}{2\pi r}\).

Step 1: Define Current Density and Total Current

The current density is given by the formula \(J = ar\), where \(a\) is a constant and \(r\) is the distance from the center of the wire. Since the current density varies linearly with \(r\), the total current \(I\) can be found by integrating \(J\) over the cross-sectional area of the wire.

Step 2: Set Up the Integral for Total Current

The total current \(I\) through the wire is obtained by integrating the current density over the cross-sectional area: \[ I = \int_0^R J \, dA = \int_0^R ar \, dA \] where \(dA = 2\pi r \, dr \) is the differential area element in cylindrical coordinates.

Step 3: Solve the Integral

Substitute \(dA = 2\pi r \, dr \) into the integral: \[ I = \int_0^R ar (2\pi r \, dr) = 2\pi a \int_0^R r^2 \, dr \] Evaluating the integral gives: \[ 2\pi a \left[ \frac{r^3}{3} \right]_0^R = \frac{2\pi a R^3}{3} \] Set this equal to the total current \(I\).

Step 4: Solve for Constant a

Equate the integral result to the total current \(I\): \[ \frac{2\pi a R^3}{3} = I \] Solving for \(a\), we find: \[ a = \frac{3I}{2\pi R^3} \] This is the expression for \(a\) in terms of \(I\) and \(R\).

Step 5: Apply Ampere's Law for r ≤ R

Ampere's law states that the line integral of the magnetic field \(B\) around a closed loop is equal to the current enclosed by the loop. For \(r \leq R\), all current within radius \(r\) contributes to the magnetic field: \[ B(2\pi r) = \oint B \, ds = \int_0^r Jr \, 2\pi r \, dr = 2\pi a \int_0^r r^2 \, dr \] Upon evaluating, we find: \[ 2\pi a \left[ \frac{r^3}{3} \right]_0^r = \frac{2\pi a r^3}{3} \] Substitute for \(a\) and solve for \(B\): \[ B=\frac{a r^2}{2} = \frac{3Ir^2}{4\pi R^3} \]

Step 6: Calculate Magnetic Field for r ≥ R

For \(r \geq R\), the total current enclosed is the entire current \(I\). Using Ampere's law, \(B(2\pi r) = I\). Therefore, \[ B = \frac{I}{2\pi r} \] This gives us the magnetic field outside the wire.

Key Concepts

Magnetic Field

A magnetic field is an invisible entity produced by electric currents or changes in electric fields. It can be visualized as a series of lines indicating the force direction that acts on moving charges or magnetic materials. When a current flows through a wire, it generates a circular magnetic field around the wire. The strength and direction of this field depend on factors such as the current's magnitude and distribution across the wire.In our example, you have a wire with a current density that varies along the radius as described by the formula \(J = ar\). This variation means the magnetic field will also differ at various points around and outside the wire. For calculations, this scenario requires careful application of Ampere's Law. By understanding the magnetic field configuration, you can predict how the magnetic field interacts with surrounding materials and other fields.

Ampere's Law

Ampere's Law is a fundamental principle used to relate magnetic fields to the electric currents that produce them. It states that the closed-loop line integral of the magnetic field \(B\) around a loop is directly proportional to the electric current passing through that loop.This principle is expressed mathematically as:\[ \oint B \cdot ds = \mu_0 I_{enc} \]where \(\mu_0\) is the permeability of free space, and \(I_{enc}\) is the current enclosed by the loop. In practice, when solving problems related to wires carrying current, like in our exercise, Ampere's Law helps determine the magnitude and direction of the magnetic field at a distance \(r\) from the wire.In our exercise, for \(r \leq R\), the current enclosed is decided by integrating the current density \(J = ar\) over the loop's area. For \(r \geq R\), the total current is \(I\), affecting the formula and calculation accordingly. Understanding Ampere's Law and how to apply it is crucial for determining the magnetic fields in such scenarios.

Integration in Cylindrical Coordinates

When dealing with problems involving cylindrical objects like wires, it's often most efficient to use cylindrical coordinates for integration. These coordinates consider radial distance \(r\), angle \(\theta\), and height \(z\) in a three-dimensional space.Cylindrical coordinates are especially useful when objects possess cylindrical symmetry, enabling simpler integration by exploiting symmetry properties. The problem from the exercise makes use of this system by evaluating the current density over a circular cross-section.In cylindrical coordinates, the differential area element \(dA\) is expressed as \(2\pi r \cdot dr\), which calculates a slice of the cylinder. By integrating current density \(J = ar\) over this element, you can find the total current by treating the wire's cross-section as a collection of infinitesimally small concentric rings.Understanding how to set up integrals in cylindrical coordinates is key for students working on electromagnetism problems, enabling them to more efficiently solve for quantities like total current or magnetic field strengths around wires.