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Chapter 21: Problem 72

Two long, straight wires are separated by \(0.120 \mathrm{~m}\). The wires carry currents of \(8.0 \mathrm{~A}\) in opposite directions, as the drawing indicates. Find the magnitude of the net magnetic field at the points labeled (a) \(A\) and (b) \(B\).

Short Answer

Expert verified
Net B at A is 0 T; net B at B is \(2.66 \times 10^{-5} \mathrm{~T}\).

Step by step solution

01

Understanding the Problem

We have two long, straight wires separated by a distance of 0.120 m. Each wire carries a current of 8.0 A in opposite directions. We need to find the net magnetic field at two points, A and B, near these wires.
02

Identifying Relevant Formula

The magnetic field produced by a long straight wire at a distance \( r \) from it is given by the formula \( B = \frac{\mu_0 I}{2\pi r} \), where \( \mu_0 = 4\pi \times 10^{-7} \mathrm{~T\cdot m/A} \) is the permeability of free space and \( I \) is the current.
03

Calculating Magnetic Field at Point A

Point A is aligned with one of the wires. The magnetic field from a wire (due to the current) at this point will be calculated using the formula \( B = \frac{\mu_0 I}{2\pi r} \). Each wire contributes a magnetic field of equal magnitude but in opposite directions at point A.
04

Substituting Values for Point A

Let \( r = 0.120 \mathrm{~m} \). For each wire, \( B = \frac{4\pi \times 10^{-7} \times 8.0}{2\pi \times 0.120} = \frac{32 \times 10^{-7}}{0.240} = 1.33 \times 10^{-5} \mathrm{~T} \). Since the currents are in opposite directions, the net magnetic field at A is zero.
05

Calculating Magnetic Field at Point B

Point B is equidistant from both wires, so the contribution of the magnetic field at this point from each wire must be added. However, since one current flows up and the other flows down, the directions of the magnetic fields at point B due to each wire are such that they add up.
06

Substituting Values for Point B

For each wire at B, \( B = \frac{4\pi \times 10^{-7} \times 8.0}{2\pi \times 0.120} = 1.33 \times 10^{-5} \mathrm{~T} \). The contributions are in the same direction, so they add up: \( B_{net,B} = 2 \times 1.33 \times 10^{-5} \mathrm{~T} = 2.66 \times 10^{-5} \mathrm{~T} \).

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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 in electromagnetism that helps us understand how magnetic fields are generated by electric currents. According to this law, the magnetic field around a closed loop is proportional to the total electric current flowing through that loop.
It is mathematically expressed as: \[ \oint B \cdot dl = \mu_0 I \]where:
  • \( B \) is the magnetic field
  • \( dl \) is a differential length element of a closed path
  • \( \mu_0 \) is the permeability of free space
  • \( I \) is the current enclosed by the path
In the context of our exercise, Amp猫re's law allows us to understand the magnetic field generated by each wire due to their currents. Since the wires carry current in opposite directions, Amp猫re's law helps us visualize how the resulting magnetic fields will interact either to cancel each other out or combine.
permeability of free space
The permeability of free space, denoted as \( \mu_0 \), is a physical constant important in the field of electromagnetism. It describes how a magnetic field can propagate through the vacuum of space. The constant is defined as: \[ \mu_0 = 4\pi \times 10^{-7} \, \mathrm{T\cdot m/A} \]Here, \( \mathrm{T} \) stands for Tesla, which is a unit of the magnetic field strength, \( \mathrm{m} \) for meters, and \( \mathrm{A} \) for Amperes, the unit of current.
This constant essentially sets the scale for how strong the magnetic field is for a given current in free space. In the context of the exercise, \( \mu_0 \) is used in the equation to calculate the magnitude of the magnetic field generated by each wire at a specific distance \( r \), ensuring our calculations account for how effectively the magnetic field propagates through space.
current in wires
Current in wires is crucial when studying magnetic fields because it's the flow of electric charges that produces these fields. In our problem, each wire carries a current of \( 8.0 \mathrm{~A} \). The direction of this current significantly affects the resulting magnetic field around the wires:
  • The magnetic field direction depends on the current direction according to the right-hand rule.
  • Opposite current directions in nearby wires can result in complex interactions between the fields.
In the exercise, these principles demonstrate that although the wires carry equal currents, their opposite directions lead to distinct magnetic field interactions at points A and B. At point A, the fields cancel each other, resulting in a net field of zero. At point B, however, they add up to give a non-zero net magnetic field.

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Most popular questions from this chapter

A copper rod of length \(0.85 \mathrm{~m}\) is lying on a frictionless table (see the drawing). Each end of the rod is attached to a fixed wire by an unstretched spring that has a spring constant of \(k=75 \mathrm{~N} / \mathrm{m}\). A magnetic field with a strength of \(0.16 \mathrm{~T}\) is oriented perpendicular to the surface of the table. (a) What must be the direction of the current in the copper rod that causes the springs to stretch? (b) If the current is \(12 \mathrm{~A}\), by how much does each spring stretch?

The maximum torque experienced by a coil in a 0.75 -T magnetic field is \(8.4 \times 10^{-4} \mathrm{~N} \cdot \mathrm{m}\). The coil is circular and consists of only one turn. The current in the coil is \(3.7 \mathrm{~A}\). What is the length of the wire from which the coil is made?

An \(\alpha\) -particle has a charge of \(+2 e\) and a mass of \(6.64 \times 10^{-27} \mathrm{~kg} .\) It is accelerated from rest through a potential difference that has a value of \(1.20 \times 10^{6} \mathrm{~V}\) and then enters a uniform magnetic field whose magnitude is \(2.20 \mathrm{~T}\). The \(\alpha\) -particle moves perpendicular to the magnetic field at all times. What is (a) the speed of the \(\alpha\) -particle, (b) the magnitude of the magnetic force on it, and (c) the radius of its circular path?

Each of these problems consists of Concept Questions followed by a related quantitative Problem. The Concept Questions involve little or no mathematics. They focus on the concepts with which the problems deal. Recognizing the concepts is the essential initial step in any problem-solving technique. Concept Questions A particle has a charge \(q\) and is located at the coordinate origin. An electric field \(E_{x}\) points along the \(+x\) axis. A magnetic field also exists, and its components \(B_{x}\) and \(B_{y}\) point along the \(+x\) and \(+y\) axis, respectively (see the drawing). Which, if any, of the fields \(\left(E_{x}, B_{x}, B_{y}\right)\) exert a force on the particle if it is (a) stationary and (b) moving along the \(+x\) axis? Justify your answers. (c) None of the three fields points along the \(+z\) axis. If the particle is moving along this axis, does it experience a force due to any of the fields? Provide a reason for your answer. Problem The electric and magnetic fields in the drawing are, respectively, \(E_{x}=+245 \mathrm{~N} /\)C, \(B_{x}=+1.80 \mathrm{~T},\) and \(B_{y}=+1.40 \mathrm{~T},\) and the charge is \(q=+5.60 \mu\mathrm{C}\) Calculate the force (magnitude and direction) exerted on the particle by each of the three fields when it is (a) stationary, (b) moving along the \(+x\) axis at a speed of \(375 \mathrm{~m} / \mathrm{s},\) and \((\mathrm{c})\) moving along the \(+z\) axis at a speed of \(375 \mathrm{~m} / \mathrm{s} .\) Be sure your answers are consistent with those in the Concept Questions.

A wire of length \(0.655 \mathrm{~m}\) carries a current of \(21.0 \mathrm{~A}\). In the presence of a \(0.470-\mathrm{T}\) magnetic field, the wire experiences a force of \(5.46 \mathrm{~N}\). What is the angle (less than \(90^{\circ}\) ) between the wire and the magnetic field?
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