The wavelength and direction of the propagating wave.

Question

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Answer 1

Question

Chapter 30, Problem 46P

(a)

To determine

The wavelength and direction of the propagating wave.

(a)

Expert Solution

Answer to Problem 46P

The wavelength of the wave is 3.0 m and direction of propagation is perpendicular to direction of magnetic field.

Explanation of Solution

Given:

The frequency is f=100 MHz .

The magnetic is B→(z,t)=(1.00×10−8 T)cos(kz−ωt)i^

Formula used:

The expression for wavelength is given by,

λ=cf

Calculation:

The wavelength of the wave is calculated as,

λ=cf=3.0× 108 m/s( ( 100 MHz )( 10 6 Hz 1 MHz ))=3.0 m

Conclusion:

Therefore, the wavelength of the wave is 3.0 m and direction of propagation is perpendicular to direction of magnetic field.

(b)

To determine

The electric field vector.

(b)

Expert Solution

Answer to Problem 46P

The electric field vector is −(3.0 V/mcos[( 2.09/ sec)z^−( 6.28× 10 8 / sec)t]j^) .

Explanation of Solution

Formula used:

The expression for electric field vector is given by,

E→(z,t)=−(Ecos[kz^−ωt]j^)

The expression for amplitude of electric field is given by,

E=cB

The expression for angular frequency is given by,

ω=2πf

The expression for wave number is given by,

k=2πλ

Calculation:

The amplitude of electric field is calculated as,

E=cB=(3.0× 108 m/s)(( 1.00× 10 −8 T))=3.0 V/m

The angular frequency is calculated as,

ω=2πf=2π(100 MHz)( 10 6 Hz 1 MHz)=6.28×108/sec

The wave number is calculated as,

k=2πλ=2π3.0 m=2.09/m

The electric field vector is calculated as,

E→(z,t)=−(Ecos[k z ^−ωt]j^)=−(3.0 V/mcos[( 2.09/ sec ) z ^−( 6.28× 10 8 / sec )t]j^)

Conclusion:

Therefore, the electric field vector is −(3.0 V/mcos[( 2.09/ sec)z^−( 6.28× 10 8 / sec)t]j^) .

(c)

To determine

The Poynting vector and intensity of the wave.

(c)

Expert Solution

Answer to Problem 46P

The Poynting vector is ((23.9 mW/ m 2))cos2[(2.09/sec)z^−(6.28× 108/sec)t](k^) and intensity of the wave is 11.9 mW/m2 .

Explanation of Solution

Formula used:

The expression for Poynting vector is given by,

S→(z,t)=1μ0E→×B→

The expression for intensity is given by,

I=|S→|

Calculation:

The Poynting vector is calculated as,

S→(z,t)=1μ0E→×B→=14π× 10 −7 N/ A 2[( −( 3.0 V/m cos[ ( 2.09/ sec ) z ^ −( 6.28× 10 8 / sec )t] j ^ ))×( ( 1.00× 10 −8 T )cos( kz−ωt ) i ^ )]=( −3.0 V/m )( 10 −8 T)4π× 10 −7 N/ A 2cos2[( 2.09/ sec)z^−( 6.28× 10 8 / sec)t](j^×i^)=(( 23.9× 10 −3 W/ m 2 )( 10 3 mW 1 W ))cos2[( 2.09/ sec)z^−( 6.28× 10 8 / sec)t](k^)

Further simplify the above,

S→(z,t)=((23.9 mW/ m 2))cos2[(2.09/sec)z^−(6.28× 108/sec)t](k^)

The intensity of the wave is calculated as,

I=|S→|=|(( 23.9 mW/ m 2 ))cos2[( 2.09/ sec )z^−( 6.28× 10 8 / sec )t](k^)|=0.5(23.9 mW/ m 2)=11.9 mW/m2

Conclusion:

Therefore, the Poynting vector is ((23.9 mW/ m 2))cos2[(2.09/sec)z^−(6.28× 108/sec)t](k^) and intensity of the wave is 11.9 mW/m2 .

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Answer 2

Question

Chapter 30, Problem 46P

(a)

To determine

The wavelength and direction of the propagating wave.

(a)

Expert Solution

Answer to Problem 46P

The wavelength of the wave is 3.0 m and direction of propagation is perpendicular to direction of magnetic field.

Explanation of Solution

Given:

The frequency is f=100 MHz .

The magnetic is B→(z,t)=(1.00×10−8 T)cos(kz−ωt)i^

Formula used:

The expression for wavelength is given by,

λ=cf

Calculation:

The wavelength of the wave is calculated as,

λ=cf=3.0× 108 m/s( ( 100 MHz )( 10 6 Hz 1 MHz ))=3.0 m

Conclusion:

Therefore, the wavelength of the wave is 3.0 m and direction of propagation is perpendicular to direction of magnetic field.

(b)

To determine

The electric field vector.

(b)

Expert Solution

Answer to Problem 46P

The electric field vector is −(3.0 V/mcos[( 2.09/ sec)z^−( 6.28× 10 8 / sec)t]j^) .

Explanation of Solution

Formula used:

The expression for electric field vector is given by,

E→(z,t)=−(Ecos[kz^−ωt]j^)

The expression for amplitude of electric field is given by,

E=cB

The expression for angular frequency is given by,

ω=2πf

The expression for wave number is given by,

k=2πλ

Calculation:

The amplitude of electric field is calculated as,

E=cB=(3.0× 108 m/s)(( 1.00× 10 −8 T))=3.0 V/m

The angular frequency is calculated as,

ω=2πf=2π(100 MHz)( 10 6 Hz 1 MHz)=6.28×108/sec

The wave number is calculated as,

k=2πλ=2π3.0 m=2.09/m

The electric field vector is calculated as,

E→(z,t)=−(Ecos[k z ^−ωt]j^)=−(3.0 V/mcos[( 2.09/ sec ) z ^−( 6.28× 10 8 / sec )t]j^)

Conclusion:

Therefore, the electric field vector is −(3.0 V/mcos[( 2.09/ sec)z^−( 6.28× 10 8 / sec)t]j^) .

(c)

To determine

The Poynting vector and intensity of the wave.

(c)

Expert Solution

Answer to Problem 46P

The Poynting vector is ((23.9 mW/ m 2))cos2[(2.09/sec)z^−(6.28× 108/sec)t](k^) and intensity of the wave is 11.9 mW/m2 .

Explanation of Solution

Formula used:

The expression for Poynting vector is given by,

S→(z,t)=1μ0E→×B→

The expression for intensity is given by,

I=|S→|

Calculation:

The Poynting vector is calculated as,

S→(z,t)=1μ0E→×B→=14π× 10 −7 N/ A 2[( −( 3.0 V/m cos[ ( 2.09/ sec ) z ^ −( 6.28× 10 8 / sec )t] j ^ ))×( ( 1.00× 10 −8 T )cos( kz−ωt ) i ^ )]=( −3.0 V/m )( 10 −8 T)4π× 10 −7 N/ A 2cos2[( 2.09/ sec)z^−( 6.28× 10 8 / sec)t](j^×i^)=(( 23.9× 10 −3 W/ m 2 )( 10 3 mW 1 W ))cos2[( 2.09/ sec)z^−( 6.28× 10 8 / sec)t](k^)

Further simplify the above,

S→(z,t)=((23.9 mW/ m 2))cos2[(2.09/sec)z^−(6.28× 108/sec)t](k^)

The intensity of the wave is calculated as,

I=|S→|=|(( 23.9 mW/ m 2 ))cos2[( 2.09/ sec )z^−( 6.28× 10 8 / sec )t](k^)|=0.5(23.9 mW/ m 2)=11.9 mW/m2

Conclusion:

Therefore, the Poynting vector is ((23.9 mW/ m 2))cos2[(2.09/sec)z^−(6.28× 108/sec)t](k^) and intensity of the wave is 11.9 mW/m2 .

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Answer 3

Question

Chapter 30, Problem 46P

(a)

To determine

The wavelength and direction of the propagating wave.

(a)

Expert Solution

Answer to Problem 46P

The wavelength of the wave is 3.0 m and direction of propagation is perpendicular to direction of magnetic field.

Explanation of Solution

Given:

The frequency is f=100 MHz .

The magnetic is B→(z,t)=(1.00×10−8 T)cos(kz−ωt)i^

Formula used:

The expression for wavelength is given by,

λ=cf

Calculation:

The wavelength of the wave is calculated as,

λ=cf=3.0× 108 m/s( ( 100 MHz )( 10 6 Hz 1 MHz ))=3.0 m

Conclusion:

Therefore, the wavelength of the wave is 3.0 m and direction of propagation is perpendicular to direction of magnetic field.

(b)

To determine

The electric field vector.

(b)

Expert Solution

Answer to Problem 46P

The electric field vector is −(3.0 V/mcos[( 2.09/ sec)z^−( 6.28× 10 8 / sec)t]j^) .

Explanation of Solution

Formula used:

The expression for electric field vector is given by,

E→(z,t)=−(Ecos[kz^−ωt]j^)

The expression for amplitude of electric field is given by,

E=cB

The expression for angular frequency is given by,

ω=2πf

The expression for wave number is given by,

k=2πλ

Calculation:

The amplitude of electric field is calculated as,

E=cB=(3.0× 108 m/s)(( 1.00× 10 −8 T))=3.0 V/m

The angular frequency is calculated as,

ω=2πf=2π(100 MHz)( 10 6 Hz 1 MHz)=6.28×108/sec

The wave number is calculated as,

k=2πλ=2π3.0 m=2.09/m

The electric field vector is calculated as,

E→(z,t)=−(Ecos[k z ^−ωt]j^)=−(3.0 V/mcos[( 2.09/ sec ) z ^−( 6.28× 10 8 / sec )t]j^)

Conclusion:

Therefore, the electric field vector is −(3.0 V/mcos[( 2.09/ sec)z^−( 6.28× 10 8 / sec)t]j^) .

(c)

To determine

The Poynting vector and intensity of the wave.

(c)

Expert Solution

Answer to Problem 46P

The Poynting vector is ((23.9 mW/ m 2))cos2[(2.09/sec)z^−(6.28× 108/sec)t](k^) and intensity of the wave is 11.9 mW/m2 .

Explanation of Solution

Formula used:

The expression for Poynting vector is given by,

S→(z,t)=1μ0E→×B→

The expression for intensity is given by,

I=|S→|

Calculation:

The Poynting vector is calculated as,

S→(z,t)=1μ0E→×B→=14π× 10 −7 N/ A 2[( −( 3.0 V/m cos[ ( 2.09/ sec ) z ^ −( 6.28× 10 8 / sec )t] j ^ ))×( ( 1.00× 10 −8 T )cos( kz−ωt ) i ^ )]=( −3.0 V/m )( 10 −8 T)4π× 10 −7 N/ A 2cos2[( 2.09/ sec)z^−( 6.28× 10 8 / sec)t](j^×i^)=(( 23.9× 10 −3 W/ m 2 )( 10 3 mW 1 W ))cos2[( 2.09/ sec)z^−( 6.28× 10 8 / sec)t](k^)

Further simplify the above,

S→(z,t)=((23.9 mW/ m 2))cos2[(2.09/sec)z^−(6.28× 108/sec)t](k^)

The intensity of the wave is calculated as,

I=|S→|=|(( 23.9 mW/ m 2 ))cos2[( 2.09/ sec )z^−( 6.28× 10 8 / sec )t](k^)|=0.5(23.9 mW/ m 2)=11.9 mW/m2

Conclusion:

Therefore, the Poynting vector is ((23.9 mW/ m 2))cos2[(2.09/sec)z^−(6.28× 108/sec)t](k^) and intensity of the wave is 11.9 mW/m2 .

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Answer 4

Question

Chapter 30, Problem 46P

(a)

To determine

The wavelength and direction of the propagating wave.

(a)

Expert Solution

Answer to Problem 46P

The wavelength of the wave is 3.0 m and direction of propagation is perpendicular to direction of magnetic field.

Explanation of Solution

Given:

The frequency is f=100 MHz .

The magnetic is B→(z,t)=(1.00×10−8 T)cos(kz−ωt)i^

Formula used:

The expression for wavelength is given by,

λ=cf

Calculation:

The wavelength of the wave is calculated as,

λ=cf=3.0× 108 m/s( ( 100 MHz )( 10 6 Hz 1 MHz ))=3.0 m

Conclusion:

Therefore, the wavelength of the wave is 3.0 m and direction of propagation is perpendicular to direction of magnetic field.

(b)

To determine

The electric field vector.

(b)

Expert Solution

Answer to Problem 46P

The electric field vector is −(3.0 V/mcos[( 2.09/ sec)z^−( 6.28× 10 8 / sec)t]j^) .

Explanation of Solution

Formula used:

The expression for electric field vector is given by,

E→(z,t)=−(Ecos[kz^−ωt]j^)

The expression for amplitude of electric field is given by,

E=cB

The expression for angular frequency is given by,

ω=2πf

The expression for wave number is given by,

k=2πλ

Calculation:

The amplitude of electric field is calculated as,

E=cB=(3.0× 108 m/s)(( 1.00× 10 −8 T))=3.0 V/m

The angular frequency is calculated as,

ω=2πf=2π(100 MHz)( 10 6 Hz 1 MHz)=6.28×108/sec

The wave number is calculated as,

k=2πλ=2π3.0 m=2.09/m

The electric field vector is calculated as,

E→(z,t)=−(Ecos[k z ^−ωt]j^)=−(3.0 V/mcos[( 2.09/ sec ) z ^−( 6.28× 10 8 / sec )t]j^)

Conclusion:

Therefore, the electric field vector is −(3.0 V/mcos[( 2.09/ sec)z^−( 6.28× 10 8 / sec)t]j^) .

(c)

To determine

The Poynting vector and intensity of the wave.

(c)

Expert Solution

Answer to Problem 46P

The Poynting vector is ((23.9 mW/ m 2))cos2[(2.09/sec)z^−(6.28× 108/sec)t](k^) and intensity of the wave is 11.9 mW/m2 .

Explanation of Solution

Formula used:

The expression for Poynting vector is given by,

S→(z,t)=1μ0E→×B→

The expression for intensity is given by,

I=|S→|

Calculation:

The Poynting vector is calculated as,

S→(z,t)=1μ0E→×B→=14π× 10 −7 N/ A 2[( −( 3.0 V/m cos[ ( 2.09/ sec ) z ^ −( 6.28× 10 8 / sec )t] j ^ ))×( ( 1.00× 10 −8 T )cos( kz−ωt ) i ^ )]=( −3.0 V/m )( 10 −8 T)4π× 10 −7 N/ A 2cos2[( 2.09/ sec)z^−( 6.28× 10 8 / sec)t](j^×i^)=(( 23.9× 10 −3 W/ m 2 )( 10 3 mW 1 W ))cos2[( 2.09/ sec)z^−( 6.28× 10 8 / sec)t](k^)

Further simplify the above,

S→(z,t)=((23.9 mW/ m 2))cos2[(2.09/sec)z^−(6.28× 108/sec)t](k^)

The intensity of the wave is calculated as,

I=|S→|=|(( 23.9 mW/ m 2 ))cos2[( 2.09/ sec )z^−( 6.28× 10 8 / sec )t](k^)|=0.5(23.9 mW/ m 2)=11.9 mW/m2

Conclusion:

Therefore, the Poynting vector is ((23.9 mW/ m 2))cos2[(2.09/sec)z^−(6.28× 108/sec)t](k^) and intensity of the wave is 11.9 mW/m2 .

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Answer 5

Chapter 30, Problem 46P

(a)

To determine

The wavelength and direction of the propagating wave.

(a)

Expert Solution

Answer to Problem 46P

The wavelength of the wave is 3.0 m and direction of propagation is perpendicular to direction of magnetic field.

Explanation of Solution

Given:

The frequency is f=100 MHz .

The magnetic is B→(z,t)=(1.00×10−8 T)cos(kz−ωt)i^

Formula used:

The expression for wavelength is given by,

λ=cf

Calculation:

The wavelength of the wave is calculated as,

λ=cf=3.0× 108 m/s( ( 100 MHz )( 10 6 Hz 1 MHz ))=3.0 m

Conclusion:

Therefore, the wavelength of the wave is 3.0 m and direction of propagation is perpendicular to direction of magnetic field.

(b)

To determine

The electric field vector.

(b)

Expert Solution

Answer to Problem 46P

The electric field vector is −(3.0 V/mcos[( 2.09/ sec)z^−( 6.28× 10 8 / sec)t]j^) .

Explanation of Solution

Formula used:

The expression for electric field vector is given by,

E→(z,t)=−(Ecos[kz^−ωt]j^)

The expression for amplitude of electric field is given by,

E=cB

The expression for angular frequency is given by,

ω=2πf

The expression for wave number is given by,

k=2πλ

Calculation:

The amplitude of electric field is calculated as,

E=cB=(3.0× 108 m/s)(( 1.00× 10 −8 T))=3.0 V/m

The angular frequency is calculated as,

ω=2πf=2π(100 MHz)( 10 6 Hz 1 MHz)=6.28×108/sec

The wave number is calculated as,

k=2πλ=2π3.0 m=2.09/m

The electric field vector is calculated as,

E→(z,t)=−(Ecos[k z ^−ωt]j^)=−(3.0 V/mcos[( 2.09/ sec ) z ^−( 6.28× 10 8 / sec )t]j^)

Conclusion:

Therefore, the electric field vector is −(3.0 V/mcos[( 2.09/ sec)z^−( 6.28× 10 8 / sec)t]j^) .

(c)

To determine

The Poynting vector and intensity of the wave.

(c)

Expert Solution

Answer to Problem 46P

The Poynting vector is ((23.9 mW/ m 2))cos2[(2.09/sec)z^−(6.28× 108/sec)t](k^) and intensity of the wave is 11.9 mW/m2 .

Explanation of Solution

Formula used:

The expression for Poynting vector is given by,

S→(z,t)=1μ0E→×B→

The expression for intensity is given by,

I=|S→|

Calculation:

The Poynting vector is calculated as,

S→(z,t)=1μ0E→×B→=14π× 10 −7 N/ A 2[( −( 3.0 V/m cos[ ( 2.09/ sec ) z ^ −( 6.28× 10 8 / sec )t] j ^ ))×( ( 1.00× 10 −8 T )cos( kz−ωt ) i ^ )]=( −3.0 V/m )( 10 −8 T)4π× 10 −7 N/ A 2cos2[( 2.09/ sec)z^−( 6.28× 10 8 / sec)t](j^×i^)=(( 23.9× 10 −3 W/ m 2 )( 10 3 mW 1 W ))cos2[( 2.09/ sec)z^−( 6.28× 10 8 / sec)t](k^)

Further simplify the above,

S→(z,t)=((23.9 mW/ m 2))cos2[(2.09/sec)z^−(6.28× 108/sec)t](k^)

The intensity of the wave is calculated as,

I=|S→|=|(( 23.9 mW/ m 2 ))cos2[( 2.09/ sec )z^−( 6.28× 10 8 / sec )t](k^)|=0.5(23.9 mW/ m 2)=11.9 mW/m2

Conclusion:

Therefore, the Poynting vector is ((23.9 mW/ m 2))cos2[(2.09/sec)z^−(6.28× 108/sec)t](k^) and intensity of the wave is 11.9 mW/m2 .

Answer 6

Chapter 30, Problem 46P

(a)

To determine

The wavelength and direction of the propagating wave.

(a)

Expert Solution

Answer to Problem 46P

The wavelength of the wave is 3.0 m and direction of propagation is perpendicular to direction of magnetic field.

Explanation of Solution

Given:

The frequency is f=100 MHz .

The magnetic is B→(z,t)=(1.00×10−8 T)cos(kz−ωt)i^

Formula used:

The expression for wavelength is given by,

λ=cf

Calculation:

The wavelength of the wave is calculated as,

λ=cf=3.0× 108 m/s( ( 100 MHz )( 10 6 Hz 1 MHz ))=3.0 m

Conclusion:

Therefore, the wavelength of the wave is 3.0 m and direction of propagation is perpendicular to direction of magnetic field.

Answer 7

Chapter 30, Problem 46P

(a)

To determine

The wavelength and direction of the propagating wave.

Answer 8

(a)

Expert Solution

Answer to Problem 46P

The wavelength of the wave is 3.0 m and direction of propagation is perpendicular to direction of magnetic field.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

The frequency is f=100 MHz .

The magnetic is B→(z,t)=(1.00×10−8 T)cos(kz−ωt)i^

Formula used:

The expression for wavelength is given by,

λ=cf

Calculation:

The wavelength of the wave is calculated as,

λ=cf=3.0× 108 m/s( ( 100 MHz )( 10 6 Hz 1 MHz ))=3.0 m

Conclusion:

Therefore, the wavelength of the wave is 3.0 m and direction of propagation is perpendicular to direction of magnetic field.

Answer 13

(b)

To determine

The electric field vector.

(b)

Expert Solution

Answer to Problem 46P

The electric field vector is −(3.0 V/mcos[( 2.09/ sec)z^−( 6.28× 10 8 / sec)t]j^) .

Explanation of Solution

Formula used:

The expression for electric field vector is given by,

E→(z,t)=−(Ecos[kz^−ωt]j^)

The expression for amplitude of electric field is given by,

E=cB

The expression for angular frequency is given by,

ω=2πf

The expression for wave number is given by,

k=2πλ

Calculation:

The amplitude of electric field is calculated as,

E=cB=(3.0× 108 m/s)(( 1.00× 10 −8 T))=3.0 V/m

The angular frequency is calculated as,

ω=2πf=2π(100 MHz)( 10 6 Hz 1 MHz)=6.28×108/sec

The wave number is calculated as,

k=2πλ=2π3.0 m=2.09/m

The electric field vector is calculated as,

E→(z,t)=−(Ecos[k z ^−ωt]j^)=−(3.0 V/mcos[( 2.09/ sec ) z ^−( 6.28× 10 8 / sec )t]j^)

Conclusion:

Therefore, the electric field vector is −(3.0 V/mcos[( 2.09/ sec)z^−( 6.28× 10 8 / sec)t]j^) .

Answer 14

(b)

To determine

The electric field vector.

Answer 15

(b)

Expert Solution

Answer to Problem 46P

The electric field vector is −(3.0 V/mcos[( 2.09/ sec)z^−( 6.28× 10 8 / sec)t]j^) .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Formula used:

The expression for electric field vector is given by,

E→(z,t)=−(Ecos[kz^−ωt]j^)

The expression for amplitude of electric field is given by,

E=cB

The expression for angular frequency is given by,

ω=2πf

The expression for wave number is given by,

k=2πλ

Calculation:

The amplitude of electric field is calculated as,

E=cB=(3.0× 108 m/s)(( 1.00× 10 −8 T))=3.0 V/m

The angular frequency is calculated as,

ω=2πf=2π(100 MHz)( 10 6 Hz 1 MHz)=6.28×108/sec

The wave number is calculated as,

k=2πλ=2π3.0 m=2.09/m

The electric field vector is calculated as,

E→(z,t)=−(Ecos[k z ^−ωt]j^)=−(3.0 V/mcos[( 2.09/ sec ) z ^−( 6.28× 10 8 / sec )t]j^)

Conclusion:

Therefore, the electric field vector is −(3.0 V/mcos[( 2.09/ sec)z^−( 6.28× 10 8 / sec)t]j^) .

Answer 20

(c)

To determine

The Poynting vector and intensity of the wave.

(c)

Expert Solution

Answer to Problem 46P

The Poynting vector is ((23.9 mW/ m 2))cos2[(2.09/sec)z^−(6.28× 108/sec)t](k^) and intensity of the wave is 11.9 mW/m2 .

Explanation of Solution

Formula used:

The expression for Poynting vector is given by,

S→(z,t)=1μ0E→×B→

The expression for intensity is given by,

I=|S→|

Calculation:

The Poynting vector is calculated as,

S→(z,t)=1μ0E→×B→=14π× 10 −7 N/ A 2[( −( 3.0 V/m cos[ ( 2.09/ sec ) z ^ −( 6.28× 10 8 / sec )t] j ^ ))×( ( 1.00× 10 −8 T )cos( kz−ωt ) i ^ )]=( −3.0 V/m )( 10 −8 T)4π× 10 −7 N/ A 2cos2[( 2.09/ sec)z^−( 6.28× 10 8 / sec)t](j^×i^)=(( 23.9× 10 −3 W/ m 2 )( 10 3 mW 1 W ))cos2[( 2.09/ sec)z^−( 6.28× 10 8 / sec)t](k^)

Further simplify the above,

S→(z,t)=((23.9 mW/ m 2))cos2[(2.09/sec)z^−(6.28× 108/sec)t](k^)

The intensity of the wave is calculated as,

I=|S→|=|(( 23.9 mW/ m 2 ))cos2[( 2.09/ sec )z^−( 6.28× 10 8 / sec )t](k^)|=0.5(23.9 mW/ m 2)=11.9 mW/m2

Conclusion:

Therefore, the Poynting vector is ((23.9 mW/ m 2))cos2[(2.09/sec)z^−(6.28× 108/sec)t](k^) and intensity of the wave is 11.9 mW/m2 .

Answer 21

(c)

To determine

The Poynting vector and intensity of the wave.

Answer 22

(c)

Expert Solution

Answer to Problem 46P

The Poynting vector is ((23.9 mW/ m 2))cos2[(2.09/sec)z^−(6.28× 108/sec)t](k^) and intensity of the wave is 11.9 mW/m2 .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Formula used:

The expression for Poynting vector is given by,

S→(z,t)=1μ0E→×B→

The expression for intensity is given by,

I=|S→|

Calculation:

The Poynting vector is calculated as,

S→(z,t)=1μ0E→×B→=14π× 10 −7 N/ A 2[( −( 3.0 V/m cos[ ( 2.09/ sec ) z ^ −( 6.28× 10 8 / sec )t] j ^ ))×( ( 1.00× 10 −8 T )cos( kz−ωt ) i ^ )]=( −3.0 V/m )( 10 −8 T)4π× 10 −7 N/ A 2cos2[( 2.09/ sec)z^−( 6.28× 10 8 / sec)t](j^×i^)=(( 23.9× 10 −3 W/ m 2 )( 10 3 mW 1 W ))cos2[( 2.09/ sec)z^−( 6.28× 10 8 / sec)t](k^)