/*! 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} Q15E (a) A long, straight solenoid ha... [FREE SOLUTION] | 91影视

91影视

(a) A long, straight solenoid has N turns, uniform cross-sectional area A, and length l. Show that the inductance of this solenoid is given by the equation L=0AN2l. Assume that the magnetic field is uniform inside the solenoid and zero outside. (Your answer is approximate because B is actually smaller at the ends than at the center. For this reason, your answer is actually an upper limit on the inductance.) (b) A metallic laboratory spring is typically 5cmlong and 0.15cmin diameter and has 50coils. If you connect such a spring in an electric circuit, how much self-inductance must you include for it if you model it as an ideal solenoid?

Short Answer

Expert verified

a)L=0AN2l

b)role="math" localid="1664194101205" L=1.1110-7H

Step by step solution

01

Magnetic field in a solenoid

For a long solenoid ofNturns, lengthiand currenti, the magnetic field is given byB=0ni=0NIi.

For the area of cross-section A, the magnetic flux is given by role="math" localid="1664194375207" =BA=0NAil.

The self-inductance is defined as L=NBi.

Clearly, after substituting the flux equation, we will getL=0N2Al .

02

Calculate the self-inductance

Given that I=5cmand D=.015cmandN=50.

role="math" localid="1664194642248" A=D24A=0.001524A=1.76710-6m2

The self-inductance is given by

L=410-75021.76710-60.05L=1.1110-7H

So, the self-inductance is1.1110-7H.

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

A cylindrical rod has resistance R. If we triple its length and diameter, what is its resistance in terms of R

An electron moves at 1.40106m/sthrough a regionin which there is a magnetic field of unspecified direction and magnitude 7.4010-2T. (a) What are the largest and smallest possible magnitudes of the acceleration of the electron due to the magnetic field? (b) If the actual acceleration of the electron is one-fourth of the largest magnitude in part (a), what is the angle
between the electron velocity and the magnetic field?

In the circuit shown in Fig. E26.49, C = 5.90 mF, 詯 = 28.0 V, and the emf has negligible resistance. Initially, the capacitor is uncharged and the switch S is in position 1. The switch is then moved to position 2 so that the capacitor begins to charge. (a) What will be the charge on the capacitor a long time after S is moved to position 2? (b) After S has been in position 2 for 3.00 ms, the charge on the capacitor is measured to be 110 mC What is the value of the resistance R? (c) How long after S is moved to position 2 will the charge on the capacitor be equal to 99.0% of the final value found in part (a)?

Why does an electric light bulb nearly always burn out just as you turn on the light, almost never while the light is shining?

An electrical conductor designed to carry large currents has a circular cross section 2.50 mm in diameter and is 14.0 m long. The resistance between its ends is 0.104. (a) What is the resistivity of the material? (b) If the electric-field magnitude in the conductor is 1.28 V/m, what is the total current? (c) If the material has 8.51028free electrons per cubic meter, find the average drift speed under the conditions of part (b).

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.