A charge q is distributed uniformly around a thin ring of radiusr . The ring is rotating about an axis through its center and perpendicular to its plane, at an angular speedω. (a) Show that the magnetic moment due to the rotating charge has magnitude μ=12qωr2. (b) What is the direction of this magnetic moment if the charge is positive?

Short Answer

Expert verified

(a) The magnitude of the magnetic moment due to the rotating chargeisμ=12qωr2

(b) The direction of the thumb of the right-hand points in the direction of dipole moment if the charge is positive and the fingers are rotating in the direction of rotation.

Step by step solution

01

 Step 1: Identification of the given data

The uniformly distributed charge over the ring is,q

The radius of the ring is, r

The angular speed of the ring is,ω

02

Representation of various formulae

The magnetic moment of the rotating charge is equal to the product of the current passing through the ring and the area of the ring. It is expressed as follows,

μ=iA

Here,i is thecurrent passing through the ring, andAis the area of the ring.

The expression for thecurrent passing through the ring is expressed as follows,

i=qt

Here,qis the charge on the object, andtis the time.

The expression for the angular speed of the ring is given as follows,

ω=2πt

Here, tis the time taken.

03

Step 3:(a) Determination of the magnitude of the magnetic moment due to the rotating charge

Substituteqtfor i, andπr2 forA inμ=iA .

μ=qt×πr2

Rearrangeω=2πt .

t=2πω

Substitute the above value in μ=qt×πr2.

μ=×πr2(2π)=12qωr2

Thus, the magnitude of the magnetic moment due to the rotating charge isμ=12qωr2

04

Step 4:(b) Determination of direction of magnetic moment if charge is positive

According to the right-hand thumb rule, if the charge is positive, on curling the fingers of right hand in the direction of rotation, the thumb points in the direction ofthedipole moment.

Thus, the direction of the thumb of the right-hand points in the direction of dipole moment if the charge is positive and the fingers are rotating in the direction of rotation.

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 Vaia!

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

Consider a solid containing Natoms per unit volume, each atom having a magnetic dipole moment μ. Suppose the direction ofμcan be only parallel or anti-parallel to an externally applied magnetic field (this will be the case if is due to the spin of a single electron). According to statistical mechanics, the probability of an atom being in a state with energy Uis proportional toe-UKT, where Tis the temperature and kis Boltzmann’s constant. Thus, because energy Uis, the fraction of atoms whose dipole moment is parallel to is proportional toeμBKTand the fraction of atoms whose dipole moment is anti-parallel to is proportional toe-μBKT. (a) Show that the magnitude of the magnetization of this solid isM=μNtanh(μBkT). Here tanh is the hyperbolic tangent function:tanh(x)=(ex-e-x)/(ex+e-x)(b) Show that the result given in (a) reduces toM=2B/kTforμBkT(c) Show that the result of (a) reduces torole="math" localid="1662964931865" M=forrole="math" localid="1662964946451" μBkT.(d) Show that both (b) and (c) agree qualitatively with Figure.

Earth has a magnetic dipole moment of μ=8×1022J/T. (a) What current would have to be produced in a single turn of wire extending around Earth at its geomagnetic equator if we wished to set up such a dipole? Could such an arrangement be used to cancel out Earth’s magnetism (b) at points in space well above Earth’s surface or (c) on Earth’s surface?

An electron is placed in a magnetic field B that is directed along azaxis. The energy difference between parallel and antiparallel alignments of the zcomponent of the electron’s spin magnetic moment withBis6.00×10-25J. What is the magnitude ofB?

At what rate must the potential difference between the plates of a parallel-plate capacitor with a2.0μF capacitance be changed to produce a displacement current of1.5A?

The saturation magnetization Mmax of the ferromagnetic metal nickel is4.70×105A/m . Calculate the magnetic dipole moment of a single nickel atom. (The density of nickel is8.90g/cm3, and its molar mass is58.71g/mol .)

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.

Sign-up for free