/*! 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 65 A long solenoid has 1400 turns p... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A long solenoid has 1400 turns per meter of length, and it carries a current of \(3.5 \mathrm{~A}\). A small circular coil of wire is placed in side the solenoid with the normal to the coil oriented at an angle of \(90.0^{\circ}\) with respect to the axis of the solenoid. The coil consists of 50 turns, has an area of \(1.2 \times 10^{-3} \mathrm{~m}^{2},\) and carries a current of \(0.50 \mathrm{~A}\). Find the torque exerted on the coil.

Short Answer

Expert verified
The torque exerted on the coil is approximately \(1.32 \times 10^{-4} \mathrm{~Nm} \).

Step by step solution

01

Calculate the Magnetic Field in the Solenoid

The magnetic field inside a long solenoid is given by the formula \( B = \mu_0 n I_s \), where \( \mu_0 \) is the permeability of free space \( (4\pi \times 10^{-7} \text{ Tm/A}) \), \( n \) is the number of turns per unit length \( (1400 \, \text{ turns/m}) \), and \( I_s \) is the current in the solenoid \( (3.5 \, \text{ A}) \). Plugging these values in, we have:\[B = (4\pi \times 10^{-7} \, \text{ Tm/A})(1400 \, \text{ turns/m})(3.5 \, \text{ A})\]
02

Calculate the Torque on the Coil

The torque \( \tau \) on a coil in a magnetic field is calculated using the formula \( \tau = n' I_c A B \sin \theta \), where \( n' \) is the number of turns in the coil \( (50) \), \( I_c \) is the current in the coil \( (0.50 \, \text{ A}) \), \( A \) is the area of the coil \( (1.2 \times 10^{-3} \, \text{ m}^2) \), \( B \) is the magnetic field calculated in Step 1, and \( \theta \) is the angle between the normal to the coil and the solenoid \((90^{\circ})\). Since \( \sin 90^{\circ} = 1 \), the formula simplifies to:\[\tau = 50 \times 0.50 \times 1.2 \times 10^{-3} \times B \times 1\]
03

Calculate the Final Torque Value

Substituting the magnetic field value calculated in Step 1 into the torque formula:\[\tau = 50 \times 0.50 \times 1.2 \times 10^{-3} \times (4\pi \times 10^{-7} \times 1400 \times 3.5)\]Simplifying gives the torque value \( \tau \).

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.

Solenoid Magnetic Field
A solenoid is basically a coil of wire, which generates a magnetic field when an electric current is passed through it. The strength of this magnetic field inside a long solenoid can be determined by the formula \( B = \mu_0 n I_s \). Here, \( \mu_0 \) represents the permeability of free space, a constant value \( (4\pi \times 10^{-7} \, \text{ Tm/A}) \).
The term \( n \) stands for the number of turns per unit length of the solenoid, which in this case is \( 1400 \, \text{ turns/m} \), and \( I_s \) is the current flowing through the solenoid, given as \( 3.5 \, \text{ A} \).
The magnetic field is uniform inside the solenoid and this uniformity is what makes solenoids quite useful in various applications, such as electromagnets or inductors in electronic circuits.
Ampere's Law
Ampere's Law relates the magnetic field around a closed loop to the electric current passing through the loop. It is represented mathematically as \( \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_\text{enc} \), where \( \mathbf{B} \) is the magnetic field, \( d\mathbf{l} \) is a differential element of the path, and \( I_\text{enc} \) is the current enclosed by the path.
In the context of a solenoid, Ampere's Law helps illustrate how the magnetic field inside the solenoid relates to the current passing through it. The magnetic field is directly proportional to the product of the current and the number of turns per unit length of the solenoid.
Understanding Ampere's Law is crucial when analyzing magnetic fields in solenoids since it provides a theoretical basis for calculating the field strength just by knowing the design parameters of the solenoid and the current it carries.
Physics Problem Solving
Solving physics problems, like the one involving solenoids and coils, requires a clear understanding of the formulas and concepts involved. Breaking down the problem into smaller steps makes it more manageable:
  • Identify given values and unknowns.
  • Choose suitable formulas related to magnetic fields and forces.
  • Perform calculations methodically step by step.
  • Make sure units are consistent across all calculations.
Analyzing each part of an equation and understanding its physical application can simplify complex problems. For example, when calculating torque on a coil inside a solenoid, you'd use the magnetic field value calculated earlier to find the torque using \( \tau = n' I_c A B \sin \theta \).
Practicing these problem-solving steps enhances one's ability to approach various physics challenges effectively.
Electromagnetic Induction
Electromagnetic induction refers to the generation of an electromotive force (EMF) across an electrical conductor when it is exposed to a varying magnetic field. This principle is foundational in many electrical technologies like transformers and generators.
In the problem at hand, a coil within a solenoid experiences a torque due to its orientation and the current it carries. If this solenoid's current and, consequently, its magnetic field were to change with time, it could induce an EMF in the coil, illustrating electromagnetic induction in action.
Transforming energy from a magnetic field into electrical energy forms the basis for devices that utilize electromagnetic induction. Grasping this concept helps in understanding and designing circuits where inductance and magnetic fields play a key role.

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 magnetic field has a magnitude of \(1.2 \times 10^{-3} \mathrm{~T}\), and an electric field has a magnitude of \(4.6 \times 10^{3} \mathrm{~N} / \mathrm{C}\). Both fields point in the same direction. A positive \(1.8-\mu \mathrm{C}\) charge moves at a speed of \(3.1 \times 10^{6} \mathrm{~m} / \mathrm{s}\) in a direction that is perpendicular to both fields. Determine the magnitude of the net force that acts on the charge.

Suppose that an ion source in a mass spectrometer produces doubly ionized gold ions \(\left(\mathrm{Au}^{2+}\right),\) each with a mass of \(3.27 \times 10^{-25} \mathrm{~kg} .\) The ions are accelerated from rest through a potential difference of \(1.00 \mathrm{kV}\). Then, a 0.500 -T magnetic field causes the ions to follow a circular path. Determine the radius of the path.

Two infinitely long, straight wires are parallel and separated by a distance of one meter. They carry currents in the same direction. Wire 1 carries four times the current as wire 2 does. On a line drawn perpendicular to both wires, locate the spot (relative to wire 1 ) where the net magnetic field is zero. Assume that wire 1 lies to the left of wire 2 and note that there are three regions to consider on this line: to the left of wire \(1,\) between wire 1 and wire \(2,\) and to the right of wire 2

Two isotopes of carbon, carbon- 12 and carbon- \(13,\) have masses of \(19.93 \times 10^{-27} \mathrm{~kg}\) and \(21.59 \times 10^{-27} \mathrm{~kg},\) respectively. These two isotopes are singly ionized \((+e)\) and each is given a speed of \(6.667 \times 10^{5} \mathrm{~m} / \mathrm{s}\). The ions then enter the bending region of a mass spectrometer where the magnetic field is \(0.8500 \mathrm{~T}\). Determine the spatial separation between the two isotopes after they have traveled through a half-circle.

The magnetic field produced by the solenoid in a magnetic resonance imaging (MRI) system designed for measurements on whole human bodies has a field strength of \(7.0 \mathrm{~T},\) and the current in the solenoid is \(2.0 \times 10^{2} \mathrm{~A} .\) What is the number of turns per meter of length of the solenoid? Note that the solenoid used to produce the magnetic field in this type of system has a length that is not very long compared to its diameter. Because of this and other design considerations, your answer will be only an approximation.

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.