/*! 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 38 A solenoid having 165 turns and ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 20: Problem 38

A solenoid having 165 turns and a cross-sectional area of 6.75 \(\mathrm{cm}^{2}\) carries a current of 1.20 A. If it is placed in a uniform 1.12 T magnetic field, find the torque this field exerts on the solenoid if its axis is oriented (a) perpendicular to the field, (b) parallel to the field, (c) at 35. \(0^{\circ}\) with the field.

Short Answer

Expert verified
(a) 0.148 N鈰卪; (b) 0 N鈰卪; (c) 0.085 N鈰卪.

Step by step solution

01

Understand the Problem

We need to calculate the torque exerted on a solenoid placed in a magnetic field with different orientations: perpendicular, parallel, and at an angle of 35 degrees. The given parameters are: number of turns \( N = 165 \), cross-sectional area \( A = 6.75 \times 10^{-4} \ m^2 \) (converted from \( cm^2 \)), current \( I = 1.20 \ A \), and magnetic field \( B = 1.12 \ T \).
02

Torque Formula for Solenoid

The formula for the torque \( \tau \) on a solenoid in a magnetic field is given by \( \tau = N \cdot I \cdot A \cdot B \cdot \sin(\theta) \), where \( \theta \) is the angle between the axis of the solenoid and the direction of the magnetic field.
03

Solve for Perpendicular Orientation (90 Degrees)

When the solenoid is perpendicular to the magnetic field, \( \theta = 90^{\circ} \) and \( \sin(90^{\circ}) = 1 \). Therefore, the torque is \( \tau = 165 \cdot 1.20 \cdot 6.75 \times 10^{-4} \cdot 1.12 \cdot 1 \). Calculating this gives \( \tau = 0.148392 \ N \cdot m \).
04

Solve for Parallel Orientation (0 Degrees)

When the solenoid is parallel to the magnetic field, \( \theta = 0^{\circ} \) and \( \sin(0^{\circ}) = 0 \). Hence, the torque is \( \tau = 165 \cdot 1.20 \cdot 6.75 \times 10^{-4} \cdot 1.12 \cdot 0 = 0 \ N \cdot m \).
05

Solve for 35 Degrees Orientation

For an angle of \( 35^{\circ} \), the sine function yields \( \sin(35^{\circ}) \approx 0.5736 \). Therefore, the torque is \( \tau = 165 \cdot 1.20 \cdot 6.75 \times 10^{-4} \cdot 1.12 \cdot 0.5736 \). Calculating this gives \( \tau \approx 0.085145 \ N \cdot m \).

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 91影视!

Key Concepts

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

Solenoid Calculations
To understand torque exerted on a solenoid, we first need to get comfortable with calculating properties of a solenoid in a magnetic field. A solenoid is a coil of wire, typically wrapped in a cylindrical shape, that produces a magnetic field when an electric current passes through it. Here are the main components necessary for solenoid calculations:
  • Number of Turns ( N ): This is the total number of loops in the coil. More turns mean a stronger magnetic field.
  • Cross-Sectional Area ( A ): This is the area of the coil's face, measured in square meters (or square centimeters that are converted into square meters). It affects the strength and reach of the magnetic field.
  • Current ( I ): The flow of electric charge through the solenoid, measured in Amperes. More current results in a stronger magnetic field.
To move forward with solenoid calculations, we translate these parameters into the magnetic characteristics of the solenoid. The next step is to consider the solenoid's orientation within a magnetic field and how it affects the resulting torque.
Magnetic Field Orientation
The orientation of the solenoid's axis relative to the magnetic field is crucial in determining the torque experienced by the solenoid. The magnetic field's strength is denoted by B, measured in Tesla. The orientation can significantly change the effects:1. **Perpendicular Orientation (90 Degrees)**: When the solenoid's axis is perpendicular to the magnetic field, this maximizes the torque. Here, the angle \( \theta = 90^{\circ} \) and hence, \( \sin(90^{\circ}) = 1 \).
2. **Parallel Orientation (0 Degrees)**: If the solenoid's axis aligns with the magnetic field, the interaction is minimized. The angle \( \theta = 0^{\circ} \) results in \( \sin(0^{\circ}) = 0 \), meaning no torque is exerted.
3. **Oblique Orientation**: Any angle other than 0 and 90 degrees needs careful calculation. For example, at \( 35^{\circ} \), the sine value is approximately 0.5736, showing that there is some torque, but not maximum.Understanding the interaction angle helps predict the solenoid's behavior effectively and calculate how the torque will change based on its orientation.
Torque Formula Application
Applying the torque formula for a solenoid in a magnetic field involves combining the concepts of N, I, A, B, and the angle \( \theta \). The torque \( \tau \) is calculated using:\[ \tau = N \cdot I \cdot A \cdot B \cdot \sin(\theta) \]This equation reflects how each variable influences the torque:
  • Number of Turns (N): More turns lead to increased torque.
  • Current (I): Higher current boosts torque.
  • Cross-Sectional Area (A): Larger area increases torque potential.
  • Magnetic Field Strength (B): Stronger magnetic fields result in more torque.
  • Angle (\( \theta \)): Torque is maximized when the solenoid is perpendicular to the field and minimized when parallel.
Calculations for different orientations (90, 0, and 35 degrees) help illustrate how each component within the formula impacts the output torque. Calculating these precisely allows for accurate predictions of how the solenoid will behave within any magnetic field setting.

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

\(\bullet\) A current in a long, straight wire produces a magnetic field of 8.0\(\mu\) t at 2.0 \(\mathrm{cm}\) from the wire's center. Answer the following questions without finding the current: (a) What is the magnetic field strength 4.0 \(\mathrm{cm}\) from the wire's center? (b) How far from the wire's center will the field be 1.0\(\mu \mathrm{T} ?\) (c) If the current were doubled, what would the field be 2.0 \(\mathrm{cm}\) from the wire's center?

\(\bullet\) A beam of protons is accelerated through a potential dif- ference of 0.745 \(\mathrm{kV}\) and then enters a uniform magnetic field traveling perpendicular to the field. (a) What magnitude of field is needed to bend these protons in a circular arc of diameter 1.75 \(\mathrm{m} ?\) (b) What magnetic field would be needed to produce a path with the same diameter if the particles were electrons having the same speed as the protons?

\(\cdot\) A particle having a mass of 0.195 g carries a charge of \(-2.50 \times 10^{-8} \mathrm{C} .\) The particle is given an initial horizontal northward velocity of \(4.00 \times 10^{4} \mathrm{m} / \mathrm{s} .\) What are the magnitude and direction of the minimum magnetic field that will balance the earth's gravitational pull on the particle?

You want to produce a magnetic field of magnitude \(5.50 \times 10^{-4} \mathrm{T}\) at a distance of 0.040 \(\mathrm{m}\) from a long, straight wire's center. (a) What current is required to produce this field? (b) With the current found in part (a), how strong is the magnetic field 8.00 \(\mathrm{cm}\) from the wire's center?

\(\bullet\) A 3.25 g bullet picks up an electric charge of 1.65\(\mu C\) as it travels down the barrel of a rifle. It leaves the barrel at a speed of 425 \(\mathrm{m} / \mathrm{s}\) , traveling perpendicular to the earth's magnetic field, which has a magnitude of \(5.50 \times 10^{-4} \mathrm{T} .\) Calculate (a) the magnitude of the magnetic force on the bullet and (b) the magnitude of the bullet's acceleration due to the magnetic force at the instant it leaves the rifle barrel.
See all solutions

Recommended explanations on Physics Textbooks

Electromagnetism

Read Explanation

Work Energy and Power

Read Explanation

Physical Quantities And Units

Read Explanation

Medical Physics

Read Explanation

Measurements

Read Explanation

Further Mechanics and Thermal 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.