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Chapter 31: Problem 13

If the \(B\) -field at a point \(P\) some distance from a straight wire in air is \(20.0 \mu \mathrm{T}\) and a current of \(20.0\) A flows in the wire, determine the perpendicular distance from the wire to point \(P\).

Short Answer

Expert verified
The perpendicular distance from the wire to point P is 2 cm.

Step by step solution

01

Understand the problem statement

We have a straight wire carrying a current of 20.0 A. The magnetic field at a point P, some distance from the wire, is 20.0 渭T. We need to find this distance.
02

Recall the formula for the magnetic field around a straight wire

The magnetic field (\( B \)) around a long straight wire is given by the formula \( B = \frac{{\mu_0 I}}{{2\pi r}} \), where \( B \) is the magnetic field strength, \( \mu_0 \) is the permeability of free space, \( I \) is the current, and \( r \) is the perpendicular distance from the wire. \( \mu_0 = 4\pi \times 10^{-7} \) Tm/A.
03

Substitute known values into the formula

We substitute the known values into the formula: \( 20.0 \times 10^{-6} = \frac{{4\pi \times 10^{-7} \times 20}}{{2\pi r}} \).
04

Simplify the equation

Simplify the equation by canceling out common terms: \( 20.0 \times 10^{-6} = \frac{{4 \times 10^{-7} \times 20}}{{2r}} \).
05

Solve for distance \( r \)

Rearrange the equation to solve for \( r \):\[ r = \frac{{4 \times 10^{-7} \times 20}}{{2 \times 20 \times 10^{-6}}} \].Calculate:\[ r = \frac{{80 \times 10^{-7}}}{{40 \times 10^{-6}}} \].Simplify this to:\[ r = 2 \times 10^{-2} \].Thus, \( r = 0.02 \, \text{m} \) or \( 2 \, \text{cm} \).

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Key Concepts

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

Magnetic Field Strength
The concept of magnetic field strength is essential when dealing with a magnetic field produced by a current-carrying wire. Magnetic field strength, denoted as \( B \), measures the intensity of a magnetic field at a given point. Its unit is Tesla (T), and it gauges how strongly the magnetic field can interact with other magnetic materials or currents. When a current flows through a wire, it generates a magnetic field around it. The strength of this field depends on the amount of current flowing through the wire and the distance from the wire. As you move further away from the wire, the magnetic field strength decreases. In scenarios involving electromagnetic applications, like the interaction between magnets and current, knowing the magnetic field strength helps us understand these forces' magnitude and direction. This knowledge is pivotal for solving various engineering and physics problems.
Permeability of Free Space
The term "permeability of free space," often represented by the symbol \( \mu_0 \), plays an important role in electromagnetic theory. It is a constant that describes how much magnetic field can pass through a vacuum. This constant is crucial for calculating magnetic fields in space free from any matter. The value of the permeability of free space is approximately \( 4\pi \times 10^{-7} \) Tm/A. This constant is used in formulas to calculate the magnetic field created by electrical currents, such as when determining the magnetic field strength around a wire. Understanding permeability helps to comprehend how magnetic fields interact in different environments. In real-world applications, permeability is used to design systems such as electromagnets, MRI machines, and other devices relying on magnetic fields. It serves as a foundation for understanding how materials respond to magnetic forces.
Current in a Wire
Current, denoted as \( I \), refers to the flow of electric charge through a conductor like a wire. Current is measured in Amperes (A), which represents the rate at which charge is flowing.When an electric current flows through a wire, it produces a magnetic field around it. This field forms concentric circles perpendicular to the wire, the direction of which can be determined using the right-hand rule. The strength and reach of this magnetic field depend on both the magnitude of the current and the distance from the wire. The more current, the stronger the magnetic field.In practical applications, understanding the relationship between current and magnetic field strength is critical for designing electrical circuits and devices. It also explains how motors and generators work, using the interaction between currents and magnetic fields to create movement or convert energy forms.
Distance from a Wire
Distance from a wire, denoted as \( r \) in magnetic calculations, is key to calculating the strength of a magnetic field created by a current-carrying wire. The magnetic field's strength decreases as the distance from the wire increases, and this is expressed in the formula for the magnetic field around a wire.The formula \( B = \frac{\mu_0 I}{2\pi r} \) includes \( r \) to represent the perpendicular distance from the wire to the point in question. To solve problems involving the magnetic field at a distance, you rearrange this formula to determine \( r \), given known values for \( B \) and \( I \). Understanding how distance affects magnetic fields is crucial in practical situations, such as determining safe operating distances around high-voltage lines or designing compact, efficient circuits. It is a fundamental concept in electromagnetic theory that impacts both theoretical physics and engineering applications.

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Most popular questions from this chapter

In Bohr's model of the hydrogen atom, the electron travels with speed \(2.2 \times 10^{6} \mathrm{~m} / \mathrm{s}\) in a circle \(\left(r=5.3 \times 10^{-11} \mathrm{~m}\right)\) about the nucleus. Find the value of \(B\) at the nucleus due to the electron's motion. Assume vacuum. In Problem \(26.17\) we found that the orbiting electron corresponds to a current loop with \(I=1.06 \mathrm{~mA}\). The field at the center of the current loop is $$B=\frac{\mu_{0} I}{2 r}=\frac{\left(4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}\right)\left(1.06 \times 10^{-3} \mathrm{~A}\right)}{2\left(5.3 \times 10^{-11} \mathrm{~m}\right)}=13 \mathrm{~T} $$

An air solenoid has 10000 turns per meter and carries a current of \(1.60 \mathrm{~A}\). What is the approximate value of the \(B\) -field at its ends when the core of the solenoid has a permeability of \(\mu=50 \mu_{0}\) ?

Two long fixed parallel wires, \(A\) and \(B\), are \(10 \mathrm{~cm}\) apart in air and carry \(40 \mathrm{~A}\) and 20 A, respectively, in opposite directions. Determine the resultant field \((a)\) on a line midway between the wires and parallel to them and \((b)\) on a line \(8.0 \mathrm{~cm}\) from wire \(A\) and \(18 \mathrm{~cm}\) from wire \(B\). (c) What is the force per meter on a third long wire, midway between \(A\) and \(B\) and in their plane, when it carries a current of \(5.0 \mathrm{~A}\) in the same direction as the current in \(A\) ?

A flat circular coil having 10 loops of wire has a diameter of \(2.0\) \(\mathrm{cm}\) and carries a current of \(0.50 \mathrm{~A}\). It is mounted inside a long solenoid immersed in air, that has 200 loops on its 25 -cm length. The current in the solenoid is \(2.4\) A. Compute the torque required to hold the coil with its central axis perpendicular to that of the solenoid. Let the subscripts \(s\) and \(c\) refer to the solenoid and coil, respectively. Then $$ \tau=N_{c} I_{c} A_{c} B_{s} \sin 90^{\circ} $$ But \(B_{s}=\mu_{0} n I_{s}=\mu_{0}\left(N_{s} / L_{s}\right) I_{s}\), which gives $$ \begin{aligned} \tau &=\frac{\mu_{0} N_{c} N_{s} I_{c} I_{s}\left(\pi r_{c}^{2}\right)}{L_{s}} \\\ &=\frac{\left(4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}\right)(10)(200)(0.50 \mathrm{~A})(2.4 \mathrm{~A}) \pi(0.010 \mathrm{~m})^{2}}{0.25 \mathrm{~m}} \\ &=3.8 \times 10^{-6} \mathrm{~N} \cdot \mathrm{m}. \end{aligned} $$

If the \(B\) -field \(1.00 \mathrm{~cm}\) from a straight wire in air is \(2.00 \mathrm{mT}\), how much current flows in the wire?
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