/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 60 A long, straight wire carrying a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A long, straight wire carrying a current of 305 A is placed in a uniform magnetic field that has a magnitude of \(7.00 \times 10^{-3} \mathrm{T}\). The wire is perpendicular to the field. Find a point in space where the net magnetic field is zero. Locate this point by specifying its perpendicular distance from the wire.

Short Answer

Expert verified
The point is \(8.71 \times 10^{-3} \mathrm{m}\) from the wire, where the fields cancel.

Step by step solution

01

Understand the Setup

We are given a long, straight wire carrying a current of 305 A. The wire is situated in a uniform magnetic field with a magnitude of \(7.00 \times 10^{-3} \mathrm{T}\). The wire is perpendicular to this field. We need to find a point in space where the net magnetic field is zero. This involves understanding the superposition of the magnetic field due to the wire and the external magnetic field.
02

Calculate the Magnetic Field due to Wire

The magnetic field produced by a long, straight wire at a distance \(r\) from it is given by Ampère's Law:\[B_{\text{wire}} = \frac{{\mu_0 I}}{{2\pi r}}\]where \(\mu_0 = 4\pi \times 10^{-7} \mathrm{T \cdot m/A}\) is the permeability of free space and \(I = 305\mathrm{A}\) is the current. We need to find \(r\) such that the net magnetic field is zero.
03

Set Net Magnetic Field to Zero

For the net magnetic field to be zero, the magnetic field due to the wire \(B_{\text{wire}}\) and the external magnetic field \(B_0\) should be equal in magnitude and opposite in direction. Therefore, we set:\[\frac{{\mu_0 I}}{{2\pi r}} = 7.00 \times 10^{-3} \mathrm{T}\]
04

Solve for Distance \(r\)

Rearrange the equation from Step 3 to solve for \(r\):\[r = \frac{{\mu_0 I}}{{2\pi \times 7.00 \times 10^{-3}}}\]Substitute \(\mu_0 = 4\pi \times 10^{-7} \mathrm{T \cdot m/A}\) and \(I = 305 \mathrm{A}\):\[r = \frac{{4\pi \times 10^{-7} \times 305}}{{2\pi \times 7.00 \times 10^{-3}}}\]Calculate \(r\):\[r \approx \frac{{1220 \times 10^{-7}}}{{14 \times 10^{-3}}} \approx 8.71 \times 10^{-3} \mathrm{m}\]
05

Identify the Required Point

The point where the net magnetic field is zero lies at a perpendicular distance of approximately \(8.71 \times 10^{-3} \mathrm{m}\) from the wire. This is where the magnetic field due to the current and the external field cancel each other out.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Magnetic Field
A magnetic field is a field that exerts forces on moving charges, such as electrons, due to their motion. This field is also what makes compass needles point north and gives magnets their distinctive ability to attract or repel. When a current flows through a wire, it generates a magnetic field around the wire. The direction of this field can be determined using the right-hand rule: if you point your thumb in the direction of the current, your fingers will curl in the direction of the magnetic field.

Magnetic fields are represented by lines that show the direction and strength of the magnetic force. These lines are closer together where the field is stronger. For a long, straight wire, the magnetic field forms concentric circles around the wire.
  • Key property: The strength of the magnetic field decreases with distance from the source (the wire).
  • Unit of measurement: The magnetic field strength is measured in Teslas (T).
Understanding magnetic fields around a current-carrying wire is essential for solving problems that involve calculating the impact of additional magnetic forces in the area, such as in the given exercise.
Ampère's Law
Ampère's Law is a fundamental principle used to determine the magnetic field created by an electric current. It relates the integrated magnetic field around a closed loop to the electric current passing through the loop. The key equation for Ampère's Law is:
\[ \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I \]
where:
  • \( \mathbf{B} \) is the magnetic field.
  • \( d\mathbf{l} \) is a differential element of the loop.
  • \( \mu_0 \) is the permeability of free space.
  • \( I \) is the current.
Ampère's Law can be particularly helpful when calculating the magnetic field in symmetrical situations, such as the magnetic field produced by a long, straight wire. For a wire, we often simplify this to:
\[ B = \frac{\mu_0 I}{2\pi r} \]
Here, \( r \) is the distance from the wire, and the equation gives the magnitude of the magnetic field at this distance.

This law helps in identifying points where magnetic fields from different sources can interact to cancel each other out, like in the context of the problem where the field of a current-carrying wire meets an external magnetic field.
Permeability of Free Space
The permeability of free space, denoted \( \mu_0 \), is a constant that represents the ability of a vacuum to support the formation of a magnetic field. It is a fundamental physical constant in electromagnetism and its value is approximately \( 4\pi \times 10^{-7} \; \mathrm{T \cdot m/A} \).

This constant is crucial when considering the strength of a magnetic field generated by a current. It appears in the formula for the magnetic field around a wire, linking the current and the field it produces.
  • \( \mu_0 \) is part of Ampère's Law, which directly relates the current through a conductor to the magnetic field it generates.
  • It ensures that the calculations of magnetic fields are consistent and accurate under the laws of physics.
Understanding \( \mu_0 \) is key for accurately solving exercises involving magnetic fields, such as finding the point where the net magnetic field around a wire and an external field equals zero. By using this constant, we can reliably calculate the magnetic influence of a current, producing precise results in context with other forces in a given system.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A particle that has an \(8.2 \mu \mathrm{C}\) charge moves with a velocity of magnitude \(5.0 \times 10^{5} \mathrm{m} / \mathrm{s}\) along the \(+x\) axis. It experiences no magnetic force, although there is a magnetic field present. The maximum possible magnetic force that the charge could experience has a magnitude of \(0.48 \mathrm{N}\). Find the magnitude and direction of the magnetic field. Note that there are two possible answers for the direction of the field.

Two coils have the same number of circular turns and carry the same current. Each rotates in a magnetic field as in Figure 21.19 . Coil 1 has a radius of \(5.0 \mathrm{cm}\) and rotates in a \(0.18-\mathrm{T}\) field. Coil 2 rotates in a \(0.42-\mathrm{T}\) field. Each coil experiences the same maximum torque. What is the radius (in \(\mathrm{cm}\) ) of coil \(2 ?\)

A long, cylindrical conductor is solid throughout and has a radius \(R\). Electric charges flow parallel to the axis of the cylinder and pass uniformly through the entire cross section. The arrangement is, in effect, a solid tube of current \(I_{0} .\) The current per unit cross-sectional area (i.e., the current density) is \(I_{0} /\left(\pi R^{2}\right) .\) Use Ampère's law to show that the magnetic field inside the conductor at a distance \(r\) from the axis is \(\mu_{0} I_{0} r /\left(2 \pi R^{2}\right)\). (Hint: For a closed path, use a circle of radius r perpendicular to and centered on the axis. Note that the current through any surface is the area of the surface times the current density.)

A small compass is held horizontally, the center of its needle a distance of \(0.280 \mathrm{m}\) directly north of a long wire that is perpendicular to the earth's surface. When there is no current in the wire, the compass needle points due north, which is the direction of the horizontal component of the earth's magnetic field at that location. This component is parallel to the earth's surface. When the current in the wire is \(25.0 \mathrm{A}\), the needle points \(23.0^{\circ}\) east of north. (a) Does the current in the wire flow toward or away from the earth's surface? (b) What is the magnitude of the horizontal component of the earth's magnetic field at the location of the compass?

The figure shows a particle that carries a charge of \(q_{0}=-2.80 \times 10^{-6} \mathrm{C} .\) It is moving along the \(+y\) axis at a speed of \(v=4.8 \times 10^{6} \mathrm{m} / \mathrm{s} .\) A magnetic field \(\overrightarrow{\mathbf{B}}\) of magnitude \(3.35 \times 10^{-5} \mathrm{T}\) is directed along the \(+z\) axis, and an electric field \(\overrightarrow{\mathbf{E}}\) of magnitude \(123 \mathrm{N} / \mathrm{C}\) points along the \(-x\) axis. Concepts: (i) What forces make up the net force acting on the particle? (ii) How do you determine the direction of the magnetic force acting on the negative charge? (iii) How do you determine the direction of the electric force acting on the negative charge? (iv) Does the fact that the charge is moving affect the values of the magnetic and electric forces? Calculations: Determine the magnitude and direction of the net force that acts on the particle.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.