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Chapter 6: Problem 26

A round wire of radius \(r_{0}\) carries a current \(I\) distributed uniformly over the cross section of the wire. Let the axis of the wire be the \(z\) axis, with \(\hat{\mathbf{z}}\) the direction of the current. Show that a vector potential of the form \(\mathbf{A}=\) constant \(\times \hat{\mathbf{z}}\left(x^{2}+y^{2}\right)\) will correctly give the magnetic field \(\mathrm{B}\) of this current at all points inside the wire. What is the value of the constant?

Short Answer

Expert verified
The constant is \( C = -\frac{\mu_0 I}{4 \pi r_0^2} \).

Step by step solution

01

Understanding the Problem

We need to find a vector potential \( \mathbf{A} \) such that its curl gives us the magnetic field \( \mathbf{B} \) inside the wire. We use the given form \( \mathbf{A} = \text{constant} \times \hat{\mathbf{z}} (x^2 + y^2) \). We will determine the constant and verify this form gives the correct \( \mathbf{B} \).
02

Express the Known Relations

The magnetic field \( \mathbf{B} \) can be expressed as \( \mathbf{B} = abla \times \mathbf{A} \). Therefore, we will compute the curl of the given vector potential \( \mathbf{A} = C \hat{\mathbf{z}} (x^2+y^2) \), where \( C \) is the constant we need to determine.
03

Calculating the Curl of \( \mathbf{A} \)

The curl of \( \mathbf{A} \) is given by \( abla \times \mathbf{A} = \left(\frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z}, \frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x}, \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y}\right) \). Since \( \mathbf{A} = C \hat{\mathbf{z}} (x^2+y^2) \), \( A_x = A_y = 0 \), and \( A_z = C(x^2+y^2) \), we get two non-zero components. Calculating these gives \( B_x = \frac{\partial A_z}{\partial y} = 2C y \) and \( B_y = -\frac{\partial A_z}{\partial x} = -2Cx \). Thus, \( \mathbf{B} = (2Cy, -2Cx, 0) \).
04

Relate Curl Result to Known Magnetic Fields

Inside a long cylindrical wire with current density \( \mathbf{J} = \frac{I}{\pi r_0^2} \hat{\mathbf{z}} \), and using Amp猫re's Law, the magnetic field at a distance \( r = \sqrt{x^2 + y^2} < r_0 \) has the magnitude \( B = \frac{\mu_0 I r}{2 \pi r_0^2} \), directed azimuthally. This corresponds to a vector \( \mathbf{B} = \left(-\frac{\mu_0 I y}{2 \pi r_0^2}, \frac{\mu_0 I x}{2 \pi r_0^2}, 0\right) \).
05

Determine the Constant

To match \( \mathbf{B} = (2Cy, -2Cx, 0) \) with the Amp猫re solution \( \mathbf{B} = \left(-\frac{\mu_0 I y}{2 \pi r_0^2}, \frac{\mu_0 I x}{2 \pi r_0^2}, 0\right) \), equate corresponding components: \(-2C = \frac{\mu_0 I}{2 \pi r_0^2}\). Thus, \( C = -\frac{\mu_0 I}{4 \pi r_0^2} \).

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Key Concepts

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

Magnetic Field
The concept of the magnetic field is central to understanding various electromagnetic phenomena. In the context of a cylindrical wire, the magnetic field is generated by the movement of electric current through the wire.
Amp猫re's Law is particularly helpful in determining this magnetic field. It relates the magnetic field in a closed loop to the electric current passing through the loop. Specifically, the magnetic field inside the wire can be determined using Amp猫re's Law, which in a cylindrical coordinate system gives:
\[ B = \frac{\mu_0 I r}{2 \pi r_0^2} \]
This formula tells us that the magnetic field strength is both dependent on the distance from the center of the wire and is directed azimuthally. Therefore, the field forms concentric circles around the axis of the wire, which is the z-axis in this case. These concentric circles represent the direction of the magnetic field lines around the wire.
Cylindrical Wire
A cylindrical wire, in the electromagnetic context, is a long, round conductor that carries an electric current. Its structure and symmetry play a crucial role in determining the distribution of the magnetic field around and within it.
The wire's axis runs along the z-direction, and the current runs parallel to this axis. Due to its cylindrical shape, it exhibits a high degree of symmetry, making calculations involving the magnetic field easier when using cylindrical coordinates.
When dealing with a cylindrical wire carrying current, we assume the current is uniformly distributed over the cross-section. This uniform distribution simplifies applying Amp猫re鈥檚 Law to compute the magnetic field. The cross-section is typically imagined as a circle of radius \( r_0 \), which represents the coverage of the current across the wire.
Current Distribution
Current distribution refers to how electric current spreads across the cross-section of a conductor. In the case of the cylindrical wire discussed, the current is assumed to be uniformly distributed, meaning that each small area unit within the cross-section carries the same amount of current density.
The uniform distribution of current is commonly expressed in terms of current density \( \mathbf{J} \), which is the current per unit area. For a cylindrical wire with current \( I \) and radius \( r_0 \), the current density is given by:
\[ \mathbf{J} = \frac{I}{\pi r_0^2} \hat{\mathbf{z}} \]
This expression indicates that the current density is constant over the circular cross-section of the wire, which significantly simplifies the calculation of the magnetic field. It also indicates the direction of the current, which aligns with the wire's axis (the z-direction), reflecting how the current flows longitudinally through the wire.

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Most popular questions from this chapter

Suppose we had a situation in which the component of the magnetic field parallel to the plane of a sheet had the same magnitude on both sides, but changed direction by \(90^{\circ}\) in going through the sheet. What is going on here? Would there be a force on the sheet? Should our formula for the force on a current sheet apply to cases like this?

For some purposes it is useful to accelerate negative hydrogen ions in a cyclotron. A negative hydrogen ion, \(\mathrm{H}^{-}\), is a hydrogen atom to which an extra electron has become attached. The attachment is fairly weak; an electric field of only \(1.5 \times 10^{4}\) statvolts \(/ \mathrm{cm}\) (a rather small field by atomic standards) will pull an electron loose, leaving a hydrogen atom. If we want to accelerate \(\mathrm{H}^{-}\) ions up to a kinetic energy of 1 Gev \(\left(10^{9}\right.\) ev), what is the highest magnetic field we dare use to keep them on a circular orbit up to final encrgy? (To find \(\gamma\) for this problem you only need the rest mass of the \(\mathrm{H}^{-}\) ion, which is of course practically the same as that of the proton, approximately 1 Gev.)

A wire carrying current \(I\) runs down the \(y\) axis to the origin, thence out to infinity along the positive \(x\) axis. Show that the magnetic field in the quadrant \(x>0, y>0\) of the \(x y\) plane is given by $$ B_{z}=\frac{I}{c}\left(\frac{1}{x}+\frac{1}{y}+\frac{x}{y \sqrt{x^{2}+y^{2}}}+\frac{y}{x \sqrt{x^{2}+y^{2}}}\right) $$

A particle of charge \(q\) and rest mass \(m\) is moving with velocity y where the magnetic field is \(\mathbf{B}\). Here \(\mathbf{B}\) is perpendicular to \(\mathbf{v}\), and there is no electric field. Show that the path of the particle is a curve with radius of curvature \(R\) given by \(R=p c / q B\), where \(p\) is the momentum of the particle, \(\beta \gamma m c .\) (Hint: Note that the force \(q \mathbf{v} \times \mathbf{B} / c\) can only change the direction of the particle's momentum, not its magnitude. By what angle \(\Delta \theta\) is the direction of \(\mathbf{p}\) changed in a short time \(\Delta t ?)\) If B is the same everywhere the particle will follow a circular path. Find the time required to complete one revolution.

Consider two solenoids, one of which is a tenth-scale model of the other. The larger solenoid is 2 meters long, and 1 meter in diameter and is wound with 1 -cm-diameter copper wire. When the coil is connected to a 120 -volt direct- current generator, the magnetic field at its center is 1000 gauss. The scaled- down model is exactly one-tenth the size in every linear dimension, including the diameter of the wire. The number of turns is the same, and it is designed to provide the same central field. (a) Show that the voltage required is the same, namely, 120 volts. (b) Compare the coils with respect to the power dissipated and the difficulty of removing this heat by some cooling means.
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