/*! 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} Q. 42 At one instant, the electric and... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

At one instant, the electric and magnetic fields at one point of an electromagnetic wave are E→=(200ı^+300ȷ^-50k^)V/mand B→=B0(7.3ı^-7.3ȷ^+ak^)μT.

a. What are the values of aand B0?

b. What is the Poynting vector at this time and position?

Short Answer

Expert verified

(a) The values of a=-14.6andBo=6.7856×10-2

(b) The Poynting vector at time and position isS→=-260i^+140j^-200k^W/m2

Step by step solution

01

Find a (part a)

By, E→⊥B→:

E→×B→=0

0=(200)(7.3)+(300)(-7.3)+(-50)a

0=1460+(-2190)-50a

50a=-730

a=-73050

a=-14.6

02

Find Bo (part a)

|E→|=c|B→|

(200)2+(300)2+(50)2=Bo23×108m/s2(7.3)2+(-7.3)2+(-14.6)2

Bo2=(200)2+(300)2+(50)23×108m/s2(7.3)2+(-7.3)2+(-14.6)2

Bo=(200)2+(300)2+(50)23×108m/s2(7.3)2+(-7.3)2+(-14.6)2

Bo=6.76×10-2

03

Find the Poynting vector

The Poynting vector is:

S→=μo-1E→×B→

S→=μo-1Bo10-6[(300)(-14.6)-(-7.3)(-50)]i^+[(-50)(7.3)-(-14.6)(200)]j^+[(200)(-7.3)-(7.3)(300)]k^

S→=μo-1Bo10-3[-4.75i^+2.56j^-3.65k^]

S→=-260i^+140j^-200k^W/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 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.