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A long, straight wire carries a current of 48 A. The magnetic field produced by this current at a certain point is \(8.0 \times 10^{-5} \mathrm{~T}\). How far is the point from the wire?

Short Answer

Expert verified
The point is 1.2 cm from the wire.

Step by step solution

01

Understanding the Formula

The magnetic field due to a long, straight wire is given by the formula \( B = \frac{\mu_0 I}{2 \pi r} \), where \( B \) is the magnetic field, \( I \) is the current, \( r \) is the distance from the wire, and \( \mu_0 \) is the permeability of free space, approximately equal to \( 4\pi \times 10^{-7} \text{ T m/A} \).
02

Rearranging the Formula

To find the distance \( r \), we rearrange the formula to solve for \( r \): \( r = \frac{\mu_0 I}{2\pi B} \).
03

Substituting Known Values

Substitute the known values into the rearranged formula: \( I = 48 \text{ A} \), \( B = 8.0 \times 10^{-5} \text{ T} \), and \( \mu_0 = 4\pi \times 10^{-7} \text{ T m/A} \). Thus, \( r = \frac{4\pi \times 10^{-7} \times 48}{2\pi \times 8.0 \times 10^{-5}} \).
04

Simplifying the Equation

Simplify the equation: the \( 2\pi \) and \( \pi \) terms cancel out partially, leading to \( r = \frac{4 \times 48 \times 10^{-7}}{2 \times 8.0 \times 10^{-5}} \).
05

Calculating the Distance

Calculate \( r \) by computing the values: \( r = \frac{192 \times 10^{-7}}{16 \times 10^{-5}} = \frac{192 \times 10^{-7}}{16} \times 10^{2} = 1.2 \times 10^{-2} \text{ m} = 0.012 \text{ m} = 1.2 \text{ cm} \).

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Key Concepts

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

Magnetic Field Strength
Magnetic field strength refers to the strength or intensity of a magnetic field at a particular point. It is denoted by the symbol \( B \) and is measured in units of Teslas (T). The strength of the magnetic field surrounding a current-carrying wire depends on several factors, primarily the amount of current flowing through the wire and the distance from the wire.

A useful formula to calculate the magnetic field strength (\( B \)) around a long, straight wire is \( B = \frac{\mu_0 I}{2 \pi r} \). In this formula:
  • \( \mu_0 \) is the permeability of free space, a constant value.
  • \( I \) is the current in amperes.
  • \( r \) is the distance from the wire.
The magnetic field decreases in strength with an increase in the distance \( r \) from the wire. Understanding how these elements interact helps in solving problems involving magnetic fields around current-carrying wires.
Ampere's Law
Ampere's Law provides a fundamental understanding of the relationship between electricity and magnetism. It states that the magnetic field around an electric current is proportional to the current. This principle forms the basis for understanding how current in a wire generates a magnetic field.

The mathematical expression of Ampere's Law is \( \oint \mathbf{B} \cdot d \mathbf{l} = \mu_0 I \), indicating the integration of the magnetic field \( \mathbf{B} \) around a closed path is equal to the permeability of free space \( \mu_0 \) times the net current \( I \) passing through any surface enclosed by the path.

Using this law in practical situations allows us to find magnetic fields arising from current distributions. It provides insight into the behavior of magnetic fields in various configurations, such as the circular fields around a straight current-carrying wire.
Permeability of Free Space
The permeability of free space, denoted by \( \mu_0 \), is a physical constant essential in electromagnetic theory. Its value is approximately \( 4\pi \times 10^{-7} \text{ T m/A} \). This constant helps relate the magnetic field in a vacuum to the current which produces it.

It's part of the formula \( B = \frac{\mu_0 I}{2 \pi r} \), playing a crucial role in determining magnetic field strength. It's important to note that \( \mu_0 \) is specific to a vacuum scenario and represents how much resistance the vacuum offers to the magnetic field lines.

Understanding \( \mu_0 \) is vital across many physics problems and applications, particularly those involving current-carrying conductors like wires. It highlights how free space affects electromagnetic principles, forming the baseline from which other materials' permeabilities are measured.
Current Electricity
Current electricity is the flow of electric charge through a conductor. It is quantified by the current, measured in amperes (A), which defines how much charge is passing through a wire per unit time.

In this specific exercise, the wire carries a current of 48 A. This higher current results in a stronger magnetic field around the wire, as seen in the exercise solution. The phenomenon is a direct application of electromagnetism principles, showing how electrified wires influence nearby magnetic fields.
  • Electric current, in physics, is represented by \( I \).
  • A steady current implies constant charge flow.
  • The direction of current (conventional current flow) is opposite to electron flow.
Comprehending current electricity provides foundational knowledge for analyzing and predicting the behavior of circuits and their effects on surrounding environments.
Distance from Wire
The distance from the wire, denoted by \( r \), is a significant factor in determining the magnetic field strength produced by a current-carrying wire. As shown in the formula \( B = \frac{\mu_0 I}{2 \pi r} \), \( r \) is inversely proportional to the magnetic field strength. This implies that the farther you move away from the wire, the weaker the magnetic field becomes.

In practical terms, when solving for \( r \) in a given exercise, you rearrange the formula to \( r = \frac{\mu_0 I}{2\pi B} \). This rearrangement allows you to directly calculate the distance from the wire when the magnetic field and current are known, providing a clear linkage between electromagnetism and spatial considerations.
  • Distance impacts the intensity of magnetic influences.
  • Predicts how field strength diminishes with increasing distance.
  • Fundamental in designing and understanding electromagnetic applications.
Understanding this concept is crucial for solving problems like the given exercise, where calculating the exact distance from a magnetic source is necessary.

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Most popular questions from this chapter

The 1200 -turn coil in a dc motor has an area per turn of \(1.1 \times 10^{-2} \mathrm{~m}^{2}\). The design for the motor specifies that the magnitude of the maximum torque is \(5.8 \mathrm{~N} \cdot \mathrm{m}\) when the coil is placed in a 0.20 -T magnetic field. What is the current in the coil?

A square coil of wire containing a single turn is placed in a uniform 0.25 -T magnetic field, as the drawing shows. Each side has a length of \(0.32 \mathrm{~m}\), and the current in the coil is 12 A. Determine the magnitude of the magnetic force on each of the four sides.

A wire carries a current of 0.66 A. This wire makes an angle of \(58^{\circ}\) with respect to a magnetic field of magnitude \(4.7 \times 10^{-5} \mathrm{~T}\). The wire experiences a magnetic force of magnitude \(7.1 \times 10^{-5} \mathrm{~N}\). What is the length of the wire?

Review Interactive Solution \(\underline{21.55} 21.55\) at for one approach to this problem. Two circular coils are concentric and lie in the same plane. The inner coil contains 140 turns of wire, has a radius of \(0.015 \mathrm{~m}\), and carries a current of \(7.2 \mathrm{~A}\). The outer coil contains 180 turns and has a radius of \(0.023 \mathrm{~m}\). What must be the magnitude and direction (relative to the current in the inner coil) of the current in the outer coil, such that the net magnetic field at the common center of the two coils is zero?

The two conducting rails in the drawing are tilted upward so they each make an angle of \(30.0^{\circ}\) with respect to the ground. The vertical magnetic field has a magnitude of \(0.050 \mathrm{~T}\) The \(0.20-\mathrm{kg}\) aluminum rod (length \(=1.6 \mathrm{~m}\) ) slides without friction down the rails at a constant velocity. How much current flows through the bar?

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