Chapter 31: Q. 42 (page 903)

At one instant, the electric and magnetic fields at one point of an electromagnetic wave are $\vec{E} = (200 \hat{ı} + 300 \hat{ȷ} - 50 \hat{k}) V / m$ and $\vec{B} = B_{0} (7.3 \hat{ı} - 7.3 \hat{ȷ} + a \hat{k}) μ T$ .
a. What are the values of $a$ and $B_{0}$ ?
b. What is the Poynting vector at this time and position?

Short Answer

Expert verified

(a) The values of $a = - 14.6$ and $B_{o} = 6.7856 \times 10^{- 2}$

(b) The Poynting vector at time and position is $\vec{S} = - 260 \hat{i} + 140 \hat{j} - 200 \hat{k} W / m^{2}$

Step by step solution

Find a (part a)

By, $\vec{E} ⊥ \vec{B}$ :

$\vec{E} \times \vec{B} = 0$

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

$0 = 1460 + (- 2190) - 50 a$

$50 a = - 730$

$a = \frac{- 730}{50}$

$a = - 14.6$

Find Bo (part a)

$| \vec{E} | = c | \vec{B} |$

$((200)^{2}) + (300)^{2} + (50)^{2} = B_{o}^{2} {(3 \times 10^{8} m / s)}^{2} ((7.3)^{2} + (- 7.3)^{2} + (- 14.6)^{2})$

$B_{o}^{2} = \frac{((200)^{2}) + (300)^{2} + ((50)^{2})}{{(3 \times 10^{8} m / s)}^{2} ((7.3)^{2} + (- 7.3)^{2} + (- 14.6)^{2})}$

$B_{o} = \sqrt{\frac{((200)^{2}) + (300)^{2} + (50)^{2}}{{(3 \times 10^{8} m / s)}^{2} ((7.3)^{2} + (- 7.3)^{2} + (- 14.6)^{2})}}$

$B_{o} = 6.76 \times 10^{- 2}$

Find the Poynting vector

The Poynting vector is:

$\vec{S} = μ_{o}^{- 1} \vec{E} \times \vec{B}$

$\vec{S} = μ_{o}^{- 1} B_{o} (10^{- 6}) \{\begin{cases} [(300) (- 14.6) - (- 7.3) (- 50)] \hat{i} \\ + [(- 50) (7.3) - (- 14.6) (200)] \hat{j} \\ + [(200) (- 7.3) - (7.3) (300)] \hat{k} \end{cases}$

$\vec{S} = μ_{o}^{- 1} B_{o} (10^{- 3}) [- 4.75 \hat{i} + 2.56 \hat{j} - 3.65 \hat{k}]$

$\vec{S} = - 260 \hat{i} + 140 \hat{j} - 200 \hat{k} W / m^{2}$

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Short Answer

Step by step solution

Find a (part a)

Find Bo (part a)

Find the Poynting vector

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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

Step by step solution

Find a (part a)

Find Bo (part a)

Find the Poynting vector

One App. One Place for Learning.

Most popular questions from this chapter

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Geometrical and Physical Optics

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Study anywhere. Anytime. Across all devices.