/*! 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} Q42E As a new electrical technician, ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Q42E (page 950)

As a new electrical technician, you are designing a large solenoid to produce a uniform 0.150-T magnetic field near the centre of the solenoid. You have enough wire for 4000 circular turns. This solenoid must be 55.0 cm long and 2.80 cm in diameter. What current will you need to produce the necessary field?

Short Answer

Expert verified

The current in the magnetic field is 16.4A

Step by step solution

01

Concept of the the magnetic field at the center of the solenoid

The magnetic field at the center of the solenoid is given by B=渭0nlwhere 渭0=4蟿蟿脳10-7H/mis permeability of free space, I is the current passing through the coil

02

STEP 2 Calculate the current in the magnetic field

To produce a uniform magnetic field of B= 0.150 T near the center of the solenoid. We have enough wire for N is 4000 circular turns and this solenoid must be L is 55.0 cm long and R is 1.40 cm in radius. The magnetic field is given asB=渭0nl where n=NLthus.B=渭0NIL

To find the current in the magnetic field substitute the values in the equation I=BL渭0Nwe have,

I=(0.150T)(0.550m)4蟺脳10鈭�7Tm/A脳4000=16.4A

Therefore, the current in the magnetic field is 16.4A. Also, the magnetic field inside the solenoid is independent of the radius of the solenoid.

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!

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

The magnetic force on a moving charged particle is always perpendicular to the magnetic fieldB鈫�. Is the trajectory of a moving charged particle always perpendicular to the magnetic field lines? Explain your reasoning.

If a 鈥�75-W鈥� bulb (see Problem 25.35) is connected across a 220-V potential difference (as is used in Europe), how much power does it dissipate? Ignore the temperature dependence of the bulb鈥檚 resistance.

Ordinary household electric lines in North America usually operate at 120 V . Why is this a desirable voltage, rather than a value considerably larger or smaller? On the other hand, automobiles usually have 12 V electrical systems. Why is this a desirable voltage?

A 10.0cm long solenoid of diameter 0.400 cm is wound uniformly with 800 turns. A second coil with 50 turns is wound around the solenoid at its center. What is the mutual inductance of the combination of the two coils?

A particle with mass1.81脳10-3kgand a charge of has1.22脳10-8C, at a given instant, a velocityV鈬赌=(3.00脳104m/s).What are the magnitude and direction of the particle鈥檚 acceleration produced by a uniform magnetic fieldB=(1.63T)i+(0.980T)j^?
See all solutions

Recommended explanations on Physics Textbooks

Particle Model of Matter

Read Explanation

Magnetism

Read Explanation

Kinematics Physics

Read Explanation

Scientific Method Physics

Read Explanation

Dynamics

Read Explanation

Physics of Motion

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.