/*! 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} Q6P Find the radiation resistance (P... [FREE SOLUTION] | 91影视

91影视

Find the radiation resistance (Prob. 11.3) for the oscillating magnetic dipole in Fig. 11.8. Express your answer in terms ofand b , and compare the radiation resistance of the electric dipole. [ Answer: 3105(b)4]

Short Answer

Expert verified

The radiation resistance for the oscillating magnetic dipole is3105b4.

Step by step solution

01

Expression for the power loss in an oscillatory magnetic dipole and the total radiated power:

Write the expression for the power loss in an oscillatory magnetic dipole.

P=I(t)2R 鈥︹ (1)

Here, I is the current, and R is the resistance.

Write the expression for the total radiated power.

<P>=0m02412c3 鈥︹ (2)

Here,0 is the magnetic constant,m0 is the magnetic dipole moment, is the angular frequency, and c is the speed of light.

02

Determine the power loss in an oscillatory magnetic dipole:

Write the expression for the current through a wire loop.

It=I0cost

Substitute It=I0costin equation (1).

P=I0cost2RP=I02Rcos2t

The average value ofcos2tis 12.

Hence, the equation for P becomes,

P=12I02R 鈥︹ (3)

03

Determine the expression for the radiation resistance:

Write the model for an oscillating magnetic dipole.

mt=b2Itz^mt=m0costz^

Here, m0=b2I0

Substitute m0=b2I0in equation (2).

P=0b2I0412c3 鈥︹ (4)

Equate equations (3) and (4).

Write the relation between angular frequency and wavelength.

=2c

Substitute =2cin equation (5).

R=02b42c46c3R=02b43c32c4R=8350cb4........(6)

04

Determine the radiation resistance for the oscillating magnetic dipole:

Substitute0=410-7H/m and c=3108m/sin equation (6).

R=835410-7H/m3108m/sb4R=3.076105b43105bR=3105b4

Therefore, the radiation resistance for the oscillating magnetic dipole is

3105b4.

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

Find the radiation resistance of the wire joining the two ends of the dipole. (This is the resistance that would give the same average power loss鈥攖o heat鈥攁s the oscillating dipole in fact puts out in the form of radiation.) Show thatR=790(d)2 , where is the wavelength of the radiation. For the wires in an ordinary radio (say, d = 5 cm ), should you worry about the radiative contribution to the total resistance?

A current I(t)flows around the circular ring in Fig. 11.8. Derive the general formula for the power radiated (analogous to Eq. 11.60), expressing your answer in terms of the magnetic dipole moment, m(t) , of the loop.

Equation 11.14 can be expressed in 鈥渃oordinate-free鈥 form by writing p0cos=p0r^. Do so, and likewise for Eqs. 11.17, 11.18. 11.19, and 11.21.

A point charge q, of mass m, is attached to a spring of constant k.Y2<<0Attimet=0it is given a kick, so its initial energy is U0=12mv02. Now it oscillates, gradually radiating away this energy.

(a) Confirm that the total energy radiated is equal to U0. Assume the radiation damping is small, so you can write the equation of motion as and the solution as

role="math" localid="1658840767865" x+y+x+02x=0,

and the solution as

x(t)=v00e-yt/2sin(0t)

with 0k/m,Y=02T, and Y2<<0 (drop Y2in comparison to 02, and when you average over a complete cycle, ignore the change in e-y).

(b) Suppose now we have two such oscillators, and we start them off with identical kicks. Regardless of their relative positions and orientations, the total energy radiated must be 2U0. But what if they are right on top of each other, so it's equivalent to a single oscillator with twice the charge; the Larmor formula says that the power radiated is four times as great, suggesting that the total will be 4U0. Find the error in this reasoning, and show that the total is actually2U0, as it should be.

An electron is released from rest and falls under the influence of gravity. In the first centimeter, what fraction of the potential energy lost is radiated away?

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.