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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 \(J=\alpha r,\) where \(\alpha\) is a constant. (a) \(\mathrm{By}\) 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 \(I\).

Short Answer

Expert verified
Therefore, the constant \(\alpha\) is given by \(\alpha = 3I/(2\pi R^3)\). The magnetic field inside the wire (for \(r ≤ R\)) is \(B = \mu_0 \frac{I}{2 \pi R^3} r^2\) and outside the wire (for \(r ≥ R\)) is \(B = \mu_0 I/(2\pi r)\).

Step by step solution

01

Work out the current density over the cross section of the wire

Given that the current density across the cross section of the wire varies as \(J = \alpha r\), the total current \(I\) is given by the integral of \(J\) over the area of the cross section. Write this as an integral over the area \(A\) of the cross section: \(I = \int_A \alpha r dA\) (1) where dA is the differential area element.
02

Decide on the form of the differential area element

dA is the area element in polar coordinates, given by \(dA = r d\theta dr\). The integral over the cross section then becomes \(I = \alpha \cdot \int_0^R \int_0^{2\pi} r^2 d\theta dr = \alpha \cdot \int_0^{2\pi} d\theta \cdot \int_0^R r^2 dr = \alpha \cdot 2\pi \cdot \int_0^R r^2 dr\).
03

Evaluate the integrals

After evaluating both integrals with respect to \(r\) and \(\theta\), we obtain \(I = \alpha \cdot 2\pi \cdot \left[ \frac{r^3}{3} \right]_{r=0}^{r=R} = \alpha \cdot 2\pi \cdot \frac{R^3}{3} = \frac{2\pi}{3} \alpha R^3\).
04

Find the constant α

Solving for \(\alpha\), we get: \(\alpha = \frac{3I}{2\pi R^3}\).
05

Start finding the magnetic field Inside the wire

Let's first consider the magnetic field inside the wire, i.e. for \(r ≤ R\). From Ampere's Law we know that \(B= \mu_0 I_{\text{enc}}/(2\pi r)\), where \(I_{\text{enc}}\) is the current enclosed by the Amperian loop. We can express this as an integral of the current density over the area enclosed by the loop, or \(I_{\text{enc}} = \int_A \alpha r dA = \alpha \cdot \int_0^r \int_0^{2\pi} r' d\theta dr'\).
06

Evaluate the Integral for Inside the wire

Performing the integral yields \(I_{\text{enc}} = \alpha \cdot 2\pi \cdot \int_0^r r'^2 dr' = \alpha \cdot 2\pi \cdot \left[ \frac{r'^3}{3} \right]_{r'=0}^{r'=r} = \frac{2\pi}{3} \alpha r^3\).
07

Use the result of α and Ampere's law for Inside the wire

Using the result for \(\alpha\) in the above equation we get \(I_{\text{enc}} = \frac{2\pi}{3} \cdot \frac{3I}{2\pi R^3} r^3 = \frac{I}{R^3} r^3\). Thus, the magnetic field inside the wire is \(B = \mu_0 \cdot \frac{I}{R^3} \cdot \frac{r^3}{2\pi r} = \mu_0 \frac{I}{2 \pi R^3} r^2\).
08

Find the Total current and Ampere's law for Outside the wire

Next, we find the magnetic field outside the wire, i.e., for \(r ≥ R\). Here the total current \(I\) is enclosed by the loop, so \(I_{\text{enc}} = I\). By Ampere's law, the magnetic field outside the wire is given by \(B = \mu_0 I_{\text{enc}}/(2\pi r) = \mu_0 I/(2\pi r)\).

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Key Concepts

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

Magnetic Field Inside a Conductor
Understanding how the magnetic field behaves within a conductor is crucial in electromagnetism. Particularly, when a steady current flows through a conductor, such as a wire with a circular cross-section, it creates a magnetic field that follows a corkscrew pattern around the wire, perpetually pointing towards the axis as per the right-hand rule.

Using Ampere's Law, which relates the magnetic field along a closed loop to the electric current passing through the loop, one can determine the magnitude of this field. Inside the conductor, the magnetic field is proportional to the distance from the center of the circular cross-section, denoted by the variable 'r'. This proportionality is because, within the wire, the 'enclosed' current contributing to the magnetic field increases with the square of 'r'. By carefully applying Ampere's Law and considering the specific current distribution, we find that the magnetic field inside the wire is given by \( B = \mu_0 \frac{I}{2 \pi R^3} r^2 \) where \(\mu_0\) is the permeability of free space, 'I' is the total current, and 'R' is the radius of the wire.
Current Density of a Wire
Current density is a measure of the electric current per unit area of cross section. In the context of a wire with a circular cross-section, we assume the current density is not uniform but varies with the distance from the center. It's given by the relation \( J = \alpha r \).

The parameter \( \alpha \) represents a constant that needs to be determined based on the total current 'I' flowing through the wire and its geometric properties. By setting up an integral equation representing the total current and solving for \( \alpha \), we determine it to be \( \alpha = \frac{3I}{2\pi R^3} \) after performing the integration in polar coordinates. Current density provides important information on how the current distributes across the wire's cross section, which in turn influences the magnetic field produced.
Circular Cross-Section Current Distribution
A wire with a circular cross-section and non-uniform current density presents an interesting distribution of current throughout its volume. Given that the current density \( J \) varies as \( \alpha r \), the current at every point within the wire is dependent on the distance from the wire's center.

This variation leads to a scenario where the outer regions of the wire carry a higher current density than the regions closer to the center. When determining the current within any subsection of this wire, one must account for how the current density changes with radius, using integration techniques over the wire's cross-sectional area. This concept is pivotal in designing electrical conductors and understanding their electromagnetic interactions.
Integration in Polar Coordinates
Integration in polar coordinates is often used when dealing with problems that have symmetrical properties, particularly those involving circular or spherical shapes. By using polar coordinates, we define any point in a plane by its distance 'r' from a reference point and the angle 'θ' with respect to a reference direction.

When solving for the current or the magnetic field inside a wire, for instance, we utilize the polar form of the area element, given by \( dA = r d\theta dr \). This allows us to integrate over circular cross-sections efficiently. For example, the current 'I' enclosed by an Amperian loop inside the conductor is found by integrating the varied current density over the loop's area, resulting in equations that include integrals such as \( \int_0^R r^2 dr \) and \( \int_0^{2\pi} d\theta \). The use of polar coordinates not only simplifies the calculation but also reflects the physical symmetry of the situation.

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Most popular questions from this chapter

Currents in dc transmission lines can be 100 A or higher. Some people are concerned that the electromagnetic fields from such lines near their homes could pose health dangers. For a line that has current \(150 \mathrm{~A}\) and a height of \(8.0 \mathrm{~m}\) above the ground, what magnetic field does the line produce at ground level? Express your answer in teslas and as a percentage of the earth's magnetic field, which is \(0.50 \mathrm{G}\). Is this value cause for worry?

A wide, long, insulating belt has a uniform positive charge per unit area \(\sigma\) on its upper surface. Rollers at each end move the belt to the right at a constant speed \(v .\) Calculate the magnitude and direction of the magnetic field produced by the moving belt at a point just above its surface. (Hint: At points near the surface and far from its edges or ends, the moving belt can be considered to be an infinite current sheet like that in Problem \(28.69 .\)

A toroidal solenoid with 400 turns of wire and a mean radius of \(6.0 \mathrm{~cm}\) carries a current of 0.25 A. The relative permeability of the core is \(80 .\) (a) What is the magnetic field in the core? (b) What part of the magnetic field is due to the magnetic moments of the atoms in the core?

A cylindrical shell with radius \(R_{1}\) and height \(H\) has charge \(Q_{1}\) and rotates around its axis with angular speed \(\omega_{1},\) as shown in Fig. \(\mathbf{P 2 8 . 7 1 .}\). Inside the cylinder, far from its edges, sits a very small disk with radius \(R_{2},\) mass \(M,\) and charge \(Q_{2}\) mounted on a pivot, spinning with a large angular velocity \(\vec{\omega}_{2}\) and oriented at angle \(\theta\) with respect to the axis of the cylinder. The center of the disk is on the axis of the cylinder. The magnetic interaction between the cylinder and the disk causes a precession of the axis of the disk. (a) What is the magnitude of the enclosed current \(I_{\text {encl }}\) surrounded by a loop that has one vertical side that is along the axis of the cylinder and extends beyond the top and bottom of the cylinder? The other vertical side of the loop is very far outside the cylinder. (b) Assume the field is uniform within the cylinder and use Ampere's law to find the magnetic field at the center of the disk. (c) The magnetic moment of the disk has magnitude \(\mu=\frac{1}{4} Q_{2} \omega_{2} R_{2}^{2} .\) What is the magnitude of the torque exerted on the disk? (d) What is the magnitude of the angular momentum of the disk?

A long, straight wire carries a current of 8.60 A. An electron is traveling in the vicinity of the wire. At the instant when the electron is \(4.50 \mathrm{~cm}\) from the wire and traveling at a speed of \(6.00 \times 10^{4} \mathrm{~m} / \mathrm{s}\) directly toward the wire, what are the magnitude and direction (relative to the direction of the current) of the force that the magnetic field of the current exerts on the electron?

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