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Chapter 28: Problem 18

Two long, straight wires, one above the other, are seperated by a distance 2\(a\) and are parallel to the \(x\) -axis. I et the \(+y\) -axis be in the plane of the wires in the direction from the lower wire to the upper wire. Each wire carries current \(I\) in the \(+x\) -direction. What are the magnitude and direction of the net magnetic field of the two wires at a point in the plane of the wires (a) midway between them; (b) at a distance \(a\) above the upper wire; \((c)\) at a distance \(a\) below the lower wire?

Short Answer

Expert verified
(a) Zero net field; (b) \(\frac{2\mu_0 I}{3\pi a}\) into the plane; (c) \(\frac{2\mu_0 I}{3\pi a}\) out of the plane.

Step by step solution

01

Understanding the Configuration

We have two parallel wires separated by a distance of 2\(a\), carrying current \(I\) in the +x-direction. The wires are positioned such that the +y-axis runs from the lower wire to the upper wire in the plane of the wires.
02

Magnetic Field Due to a Single Wire

The magnetic field due to a straight, current-carrying wire at a distance \(r\) from it is given by the formula \(B = \frac{\mu_0 I}{2\pi r}\), where \(\mu_0\) is the permeability of free space. The direction of the magnetic field follows the right-hand rule, circling around the wire.
03

Case (a): Midway Between the Wires

At a point midway between the wires, the distance from each wire to the point is \(a\). The magnetic field from each wire at this point is \(B = \frac{\mu_0 I}{2\pi a}\). One wire creates a magnetic field into the plane and the other creates a magnetic field out of the plane. These fields are equal in magnitude and opposite in direction; thus, they cancel out. The net magnetic field is zero.
04

Case (b): At a Distance a Above the Upper Wire

At a point a distance \(a\) above the upper wire, the distance to the upper wire is \(a\) and the distance to the lower wire is \(3a\). The magnetic field from the upper wire at this point is \(B_1 = \frac{\mu_0 I}{2\pi a}\) directed into the plane, and from the lower wire is \(B_2 = \frac{\mu_0 I}{6\pi a}\) also into the plane. Both fields add up, so the net magnetic field is \(B_{net} = B_1 + B_2 = \frac{\mu_0 I}{2\pi a} + \frac{\mu_0 I}{6\pi a} = \frac{2\mu_0 I}{3\pi a}\) into the plane.
05

Case (c): At a Distance a Below the Lower Wire

At a point a distance \(a\) below the lower wire, the distance to the lower wire is \(a\) and to the upper wire is \(3a\). The magnetic field from the lower wire is \(B_1 = \frac{\mu_0 I}{2\pi a}\) out of the plane, and from the upper wire is \(B_2 = \frac{\mu_0 I}{6\pi a}\) out of the plane. Both fields add up, giving a net magnetic field of \(B_{net} = B_1 + B_2 = \frac{\mu_0 I}{2\pi a} + \frac{\mu_0 I}{6\pi a} = \frac{2\mu_0 I}{3\pi a}\) out of the plane.

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Key Concepts

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

Amp猫re's Law
Amp猫re's law is a fundamental principle used to determine the magnetic field generated by a steady electric current. This law states that the magnetic field along an imaginary closed loop is proportional to the total current passing through the loop. More technically, Amp猫re's law is expressed as:\[ \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc} \]where:
  • \(\oint \vec{B} \cdot d\vec{l}\) is the line integral of the magnetic field \(\vec{B}\) around a closed loop.
  • \(\mu_0\) is the permeability of free space.
  • \(I_{enc}\) is the total current passing through the enclosed surface.
Understanding this law is crucial for figuring out the magnetic field between parallel wires as it provides the formula to calculate the magnetic field generated around a long, straight conductor carrying current. This simplifies the process of analyzing complex systems with multiple wires, as we can apply Amp猫re's law to each wire independently before considering the total magnetic field.
Superposition of Magnetic Fields
The concept of superposition is essential when we analyze the resultant magnetic field in systems with more than one current-carrying wire. Superposition implies that the total magnetic field at any point is simply the vector sum of the magnetic fields produced by each wire independently. This principle holds true because magnetic fields are vector quantities, which means both their magnitude and direction are important.When calculating the net magnetic field between the parallel wires in the exercise, the superposition principle involves:
  • Calculating the magnetic field due to each wire at the point of interest individually using the given formula \( B = \frac{\mu_0 I}{2\pi r} \).
  • Adding up the individual magnetic fields vectorially, keeping in mind their directions as given by the right-hand rule.
This means if two magnetic fields are in opposite directions, as in one of the cases discussed, they may partially or fully cancel each other out, which is why the net magnetic field results in zero or a lesser value at specific points between or beyond the wires.
Right-Hand Rule
The right-hand rule is a handy mnemonic used to easily determine the direction of a magnetic field around a current-carrying conductor. This method involves the use of your right hand and is simple to apply:
  • Point your thumb in the direction of the current flow through the wire.
  • Wrap your fingers around the wire. The direction your fingers curl is the direction of the magnetic field lines.
In the context of parallel wires, it's vital to use the right-hand rule to ensure the direction of each magnetic field is correctly identified. This helps in properly applying the superposition principle to get the net magnetic field. For example, if one wire generates a magnetic field into the plane and the other out of it, the right-hand rule assists in visualizing these directions. Thus, it directs how the field lines interact with each other, whether they add up or cancel out.

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Most popular questions from this chapter

An alpha particle (charge + 2e) and an electron move in opposite directions from the same point, each with the speed of \(2.50 \times 10^{5} \mathrm{m} / \mathrm{s}\) (Fig. \(28.32 ) .\) Find the magnitude and direction of the total magnetic field these charges produce at point \(P,\) which is 1.75 \(\mathrm{nm}\) from each of them.

A Charged Dielectric Disk. A thin disk of dielectric material with radius \(a\) has a total charge \(+Q\) distributed uniformly over its surface. It rotates \(n\) times per second about an axis perpendicular to the surface of the disk and passing through its center. Find the magnetic field at the center of the disk. (Hint: Divide the disk into concentric rings of infinitesimal width.)

As a new electrical technician, you are designing a large solenoid to produce a uniform 0.150 T magnetic field near the center of the solenoid. You have enough wire for 4000 circular turns. This solenoid must be 1.40 \(\mathrm{m}\) long and 20.0 \(\mathrm{cm}\) in diameter. What current will you need to produce the necessary field?

A long, straight, cylindrical wire of radius \(R\) carries a current uniformly distributed over its cross section. At what location is the magnetic field produced by this current equal to half of its largest value? Consider points inside and outside the wire.

A negative point charge \(q=-7.20 \mathrm{mC}\) is moving in a reference frame. When the point charge is at the origin, the magnetic field it produces at the point \(x=25.0 \mathrm{cm}, y=0, z=0\) is \(\vec{B}=(6.00 \mu \mathrm{T}) \hat{j},\) and its speed is 800 \(\mathrm{km} / \mathrm{s}\) . (a) What are the \(x\) - \(y\) .and \(z\) -components of the velocity \(\vec{v}_{0}\) of the charge? (b) At this same instant, what is the magnitude of the magnetic field that the charge produces at the point \(x=0, y=25.0 \mathrm{cm}, z=0 ?\)
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