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Chapter 20: Problem 49

A long, straight telephone cable contains six wires, each carrying a current of 0.300 A. The distances between wires can be neglected. (a) If the currents in all six wires are in the same direction, what is the magnitude of the magnetic field 2.50 \(\mathrm{m}\) from the cable? (b) If four wires carry currents in one direction and the other two carry currents in the opposite direction, what is the magnitude of the field 2.50 m from the cable?

Short Answer

Expert verified
(a) 1.44 x 10^-7 T; (b) 4.80 x 10^-8 T.

Step by step solution

01

Determine the Total Current for Part (a)

Since all six wires carry a current of 0.300 A in the same direction, the total current is the sum of these currents. Thus, the total current is given by \( I = 6 \times 0.300 \; \text{A} = 1.800 \; \text{A} \).
02

Use Ampere's Law to Find the Magnetic Field for Part (a)

The magnetic field at a distance \( r \) from a long straight conductor carrying total current \( I \) is given by Ampere's Law: \( B = \frac{\mu_0 I}{2 \pi r} \). Here, \( \mu_0 \) is the permeability of free space, \( 4\pi \times 10^{-7} \; \text{T} \cdot \text{m/A} \), and \( r = 2.50 \; \text{m} \) is the distance from the cable. Substitute the values to find \( B \):\[ B = \frac{4\pi \times 10^{-7} \times 1.800}{2 \pi \times 2.50} = \frac{4 \times 10^{-7} \times 1.800}{5} \; \text{T} = 1.44 \times 10^{-7} \; \text{T} \].
03

Determine the Net Current for Part (b)

In part (b), four wires carry currents in one direction and two in the opposite direction. Thus, their net current is \( 4 \times 0.300 - 2 \times 0.300 = 1.200 - 0.600 = 0.600 \; \text{A} \).
04

Calculate the Magnetic Field for Part (b)

Using Ampere's Law again, substitute the net current and distance into the formula \( B = \frac{\mu_0 I}{2 \pi r} \):\[ B = \frac{4\pi \times 10^{-7} \times 0.600}{2 \pi \times 2.50} = \frac{4 \times 10^{-7} \times 0.600}{5} \; \text{T} = 4.80 \times 10^{-8} \; \text{T} \].

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Key Concepts

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

Ampere's Law
To understand how we calculate the magnetic field produced by a current-carrying wire, we must familiarize ourselves with Ampere's Law. This fundamental principle of electromagnetism states that the magnetic field around a closed loop is proportional to the total current passing through the loop. Ampere's Law can be mathematically expressed as: \[ B = \frac{\mu_0 I}{2 \pi r} \]Here, \( B \) represents the magnetic field, \( \mu_0 \) is the permeability of free space, \( I \) is the total current through the loop, and \( r \) is the distance from the wire. Applying this formula helps us determine the strength of a magnetic field at any given point around the wire. This law is particularly useful when dealing with problems involving long, straight conductors.
current direction
The direction of current flow significantly influences the magnetic field's characteristics. In the exercise, we consider two scenarios: one where all currents move in the same direction and another where they move in opposite directions.
  • For part (a), all six wires carry a current in the same direction, adding up to create a stronger magnetic field.
  • For part (b), opposing currents result in partial cancellation of the magnetic field.
The current's direction is crucial because it dictates how much the fields from individual wires reinforce or oppose each other. Understanding the direction of current flow helps us correctly apply Ampere's Law and thus calculate the magnetic field efficiently.
permeability of free space
The concept of the permeability of free space, denoted as \( \mu_0 \), is essential in electromagnetism. It is a constant that quantifies the extent to which a magnetic field can penetrate vacuum. The standard value of the permeability of free space is \( 4\pi \times 10^{-7} \text{ T} \cdot \text{m/A} \).This constant appears in Ampere's Law and affects the strength of the magnetic field produced by the current. A higher permeability allows the magnetic field to be more concentrated, while a lower permeability spreads it out. When performing calculations involving magnetic fields, correctly using \( \mu_0 \) ensures accuracy in representing the interactions of fields with their environments.
total current
Calculating the total current is a straightforward step in utilizing Ampere's Law to find the magnetic field. The total current represents the sum of individual currents in a given direction within the conductor bundle.For part (a) in the exercise:
  • All six wires carry a current of 0.300 A in the same direction, resulting in a total current of \( 1.800 \text{ A} \).
In part (b):
  • The net effect of four wires carrying positive current and two carrying negative current requires subtracting the total in the opposite direction, yielding a net current of \( 0.600 \text{ A} \).
Understanding the total current aids in determining the overall influence of currents on the magnetic field. Correct calculations of total current underpin accurate assessments of magnetic interactions in diverse scenarios.

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Most popular questions from this chapter

\(\bullet \mathrm{A}^{9}\) Be nucleus containing four protons and five neutrons has a mass of \(1.50 \times 10^{-26} \mathrm{kg}\) and is traveling vertically upward at 1.35 \(\mathrm{km} / \mathrm{s} .\) If this particle suddenly enters a horizontal magnetic field of 1.12 T pointing from west to east, find the magnitude and direction of its acceleration vector the instant after it enters the field.

\(\bullet\) Atom smashers! A cyclotron particle accelerator (sometimes called an "atom smasher" in the popular press) is a device for accelerating charged particles, such as electrons and protons, to speeds close to the speed of light. The basic design is quite simple. The particle is bent in a circular path by a uniform magnetic field. An electric field is pulsed periodically to increase the speed of the particle. The charged particle (or ion) of mass \(m\) and charge \(q\) is introduced into the cyclotron so that it is moving perpendicular to a uniform magnetic field \(\vec{B}\) (a) Starting with the radius of the circular path of a charge moving in a uniform magnetic field, show that the time \(T\) for this particle to make one complete circle is \(T=\frac{2 \pi m}{|q| B}\) . (Hint: You can express the speed \(v\) in terms of \(R\) and \(T\) because the particle travels through one circumference of the circle in time \(T\) . (b) Which would take longer to complete one circle, an ion moving in a large circle or one moving in a small circle? Explain.

\(\bullet\) A singly charged ion of \(^{7} \mathrm{Li}\) (an isotope of lithium containing three protons and four neutrons) has a mass of \(1.16 \times\) \(10^{-26} \mathrm{kg} .\) It is accelerated through a potential difference of 220 \(\mathrm{V}\) and then enters a 0.723 T magnetic field perpendicular to the ion's path. What is the radius of the path of this ion in the magnetic field?

\(\bullet\) A current in a long, straight wire produces a magnetic field of 8.0\(\mu\) t at 2.0 \(\mathrm{cm}\) from the wire's center. Answer the following questions without finding the current: (a) What is the magnetic field strength 4.0 \(\mathrm{cm}\) from the wire's center? (b) How far from the wire's center will the field be 1.0\(\mu \mathrm{T} ?\) (c) If the current were doubled, what would the field be 2.0 \(\mathrm{cm}\) from the wire's center?

(a) What is the speed of a beam of electrons when the simultaneous influence of an electric field of \(1.56 \times 10^{4} \mathrm{V} / \mathrm{m}\) and a magnetic field of \(4.62 \times 10^{-3} \mathrm{T}\) , with both fields normal to the beam and to each other, produces no deflection of the electrons? (b) In a diagram, show the relative orientation of the vectors \(\vec{\boldsymbol{v}}\) , \(\vec{\boldsymbol{E}}\) and \(\vec{\boldsymbol{B}}\) . (c) When the electric field is removed, what is the radius of the electron orbit? What is the period of the orbit?
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