Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 28: Problem 48

A long solenoid (diameter of \(6.00 \mathrm{~cm}\) ) is wound with 1000 turns per meter of thin wire through which a current of 0.250 A is maintained. A wire carrying a current of 10.0 A is inserted along the axis of the solenoid. What is the magnitude of the magnetic field at a point \(1.00 \mathrm{~cm}\) from the axis?

Short Answer

Expert verified
Answer: The magnitude of the combined magnetic field at a point 1.00 cm from the axis is 0.364 T.

Step by step solution

01

Find the magnetic field due to the solenoid

To find the magnetic field inside the solenoid, we will use the formula for the magnetic field of a solenoid: \(B_{solenoid} = \mu_0 * n * I\), where \(B_{solenoid}\) is the magnetic field, \(\mu_0\) is the permeability of free space, \(n\) is the number of turns per unit length, and \(I\) is the current flowing through the solenoid. Given, \(n = 1000 \, \text{turns/m}\), \(I = 0.250 \, \text{A}\). The permeability of free space, \(\mu_0 = 4\pi \times 10^{-7} \,\text{Tm/A}\). Now, let's calculate \(B_{solenoid}\). \(B_{solenoid} = \mu_0 * n * I\)
02

Find the magnetic field due to the wire

The magnetic field generated by an infinitely long straight conductor at a distance \(r\) can be found by using the formula: \(B_{wire} = \frac{\mu_0 * I'}{2\pi r}\), where \(B_{wire}\) is the magnetic field, \(I'\) is the current in the wire, and \(r\) is the distance from the wire. Given, \(I' = 10.0 \, \text{A}\), and the distance \(r = 5.00 \, \text{cm} - 1.00 \,\text{cm} = 4.00 \,\text{cm} = 0.040 \, \text{m}\) (the radius of the solenoid minus the distance at which we want to calculate the magnetic field). Now, let's calculate \(B_{wire}\). \(B_{wire} = \frac{\mu_0 * I'}{2\pi r}\)
03

Calculate the combined magnetic field

Since both the solenoid and the wire have magnetic fields pointing in the same direction, we can add the magnitudes of these magnetic fields to get the combined magnetic field at the required point. \(B_{total} = B_{solenoid} + B_{wire}\) Now, substitute the values from Steps 1 and 2 to calculate the combined magnetic field.
04

Final calculation and answer

Now, we will substitute the given values and calculate the magnetic fields due to the solenoid and the wire, and then find the combined magnetic field. \(B_{solenoid} = (4\pi \times 10^{-7} \,\text{Tm/A}) * (1000 \, \text{turns/m}) * (0.250 \, \text{A}) = 0.314 \,\text{T}\) \(B_{wire} = \frac{(4\pi \times 10^{-7} \,\text{Tm/A}) * (10.0 \, \text{A})}{2\pi * 0.040 \, \text{m}} = 0.050 \,\text{T}\) Now, we can find the combined magnetic field: \(B_{total} = 0.314 \,\text{T} + 0.050 \,\text{T} = 0.364 \,\text{T}\) So, the magnitude of the magnetic field at a point \(1.00 \mathrm{~cm}\) from the axis is \(0.364 \,\text{T}\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Two long, straight wires are parallel to each other. The wires carry currents of different magnitudes. If the amount of current flowing in each wire is doubled, the magnitude of the force between the wires will be a) twice the magnitude of the original force. b) four times the magnitude of the original force. c) the same as the magnitude of the original force. d) half of the magnitude of the original force.

A 50-turn rectangular coil of wire of dimensions \(10.0 \mathrm{~cm}\) by \(20.0 \mathrm{~cm}\) lies in a horizontal plane, as shown in the figure. The axis of rotation of the coil is aligned north and south. It carries a current \(i=1.00 \mathrm{~A}\), and is in a magnetic field pointing from west to east. A mass of \(50.0 \mathrm{~g}\) hangs from one side of the loop. Determine the strength the magnetic field has to have to keep the loop in the horizontal orientation.

Two particles, each with charge \(q\) and mass \(m\), are traveling in a vacuum on parallel trajectories a distance \(d\) apart, both at speed \(v\) (much less than the speed of light). Calculate the ratio of the magnitude of the magnetic force that each exerts on the other to the magnitude of the electric force that each exerts on the other: \(F_{\mathrm{m}} / \mathrm{F}_{\mathrm{e}}\)

A long, straight wire lying along the \(x\) -axis carries a current, \(i\), flowing in the positive \(x\) -direction. A second long, straight wire lies along the \(y\) -axis and has a current \(i\) in the positive \(y\) -direction. What is the magnitude and the direction of the magnetic field at point \(z=b\) on the \(z\) -axis?

A wire of radius \(R\) carries current \(i\). The current density is given by \(J=J_{0}(1-r / R),\) where \(r\) is measured from the center of the wire and \(J_{0}\) is a constant. Use Ampere's Law to find the magnetic field inside the wire at a distance \(r
See all solutions

Recommended explanations on Physics Textbooks

Turning Points in Physics

Read Explanation

Dynamics

Read Explanation

Fields in Physics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Electromagnetism

Read Explanation

Medical Physics

Read Explanation
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.

Sign-up for free