/*! 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} Q15DQ Show that the units聽A.m2聽and聽... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Q15DQ (page 947)

Show that the unitsA.m2andJ/Tfor the Bohr magneton are equivalent.

Short Answer

Expert verified

They are equivalent. JT=A.m2.

Step by step solution

01

Definition of magnetic field

The term magnet field may be defined as the area around the magnet behave like amagnet.

02

Show that the units A.m2 and J/T for the Bohr magneton are equivalent.

We can show that using this

JT=N.mN/A.mJT=A.m2

Hence, they are equivalent JT=A.m2.

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 rule of thumb used to determine the internal resistance of a source is that it is the open circuit voltage divide by the short circuit current. Is this correct? Why or why not?

An idealized voltmeter is connected across the terminals of a15.0-Vbattery, and arole="math" localid="1655719696009" 75.0-惟 appliance is also connected across its terminals. If the voltmeter reads11.9V (a) how much power is being dissipated by the appliance, and (b) what is the internal resistance of the battery?

Question: A positive point charge is placed near a very large conducting plane. A professor of physics asserted that the field caused by this configuration is the same as would be obtained by removing the plane and placing a negative point charge of equal magnitude in the mirror image position behind the initial position of the plane. Is this correct? Why or why not?

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)?

Current passes through a solution of sodium chloride. In

1.00s,2.68脳1016Na+ions arrive at the negative electrode and3.92脳1016CI-

ions arrive at the positive electrode. (a) What is the current passing between

the electrodes? (b) What is the direction of the current?
See all solutions

Recommended explanations on Physics Textbooks

Work Energy and Power

Read Explanation

Famous Physicists

Read Explanation

Scientific Method Physics

Read Explanation

Modern Physics

Read Explanation

Rotational Dynamics

Read Explanation

Space 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.