The instantaneous Poynting vector.

Question

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Answer 1

Question

Chapter 30, Problem 51P

(a)

To determine

The instantaneous Poynting vector.

(a)

Expert Solution

Answer to Problem 51P

The instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

Explanation of Solution

Given:

The electric field of first wave is E→1=E10cos(k1x−ω1t)j^ .

The electric field of second wave is E→2=E20cos(k2x−ω2t+δ)j^ .

Formula used:

The expression for instantaneous Poynting vector is given by,

S→=1μ0[E→1+E→2]×1c[[E→1−E→2]i^]

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x− ω 2 t−δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x− ω 2 t−δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x− ω 2 t−δ))]i^

Conclusion:

Therefore, the instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

(b)

To determine

The time averaged Poynting vector.

(b)

Expert Solution

Answer to Problem 51P

The time averaged Poynting vector is 12μ0c[E102−E202]i^ .

Explanation of Solution

Calculation:

Consider the relation,

S→=1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t+δ))]i^

The time average of the square of the cosine term is 0.5 so, the time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the time averaged Poynting vector is 12μ0c[E102−E202]i^ .

(c)

To determine

The instantaneous and time average Poynting vector.

(c)

Expert Solution

Answer to Problem 51P

The instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

Explanation of Solution

Given:

The direction of propagation of second wave is E→2=E20cos(k2x+ω2t+δ)j^ .

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x+ ω 2 t+δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x+ ω 2 t+δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x+ ω 2 t+δ))]i^

The time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

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Answer 2

Question

Chapter 30, Problem 51P

(a)

To determine

The instantaneous Poynting vector.

(a)

Expert Solution

Answer to Problem 51P

The instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

Explanation of Solution

Given:

The electric field of first wave is E→1=E10cos(k1x−ω1t)j^ .

The electric field of second wave is E→2=E20cos(k2x−ω2t+δ)j^ .

Formula used:

The expression for instantaneous Poynting vector is given by,

S→=1μ0[E→1+E→2]×1c[[E→1−E→2]i^]

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x− ω 2 t−δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x− ω 2 t−δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x− ω 2 t−δ))]i^

Conclusion:

Therefore, the instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

(b)

To determine

The time averaged Poynting vector.

(b)

Expert Solution

Answer to Problem 51P

The time averaged Poynting vector is 12μ0c[E102−E202]i^ .

Explanation of Solution

Calculation:

Consider the relation,

S→=1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t+δ))]i^

The time average of the square of the cosine term is 0.5 so, the time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the time averaged Poynting vector is 12μ0c[E102−E202]i^ .

(c)

To determine

The instantaneous and time average Poynting vector.

(c)

Expert Solution

Answer to Problem 51P

The instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

Explanation of Solution

Given:

The direction of propagation of second wave is E→2=E20cos(k2x+ω2t+δ)j^ .

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x+ ω 2 t+δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x+ ω 2 t+δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x+ ω 2 t+δ))]i^

The time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

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Answer 3

Question

Chapter 30, Problem 51P

(a)

To determine

The instantaneous Poynting vector.

(a)

Expert Solution

Answer to Problem 51P

The instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

Explanation of Solution

Given:

The electric field of first wave is E→1=E10cos(k1x−ω1t)j^ .

The electric field of second wave is E→2=E20cos(k2x−ω2t+δ)j^ .

Formula used:

The expression for instantaneous Poynting vector is given by,

S→=1μ0[E→1+E→2]×1c[[E→1−E→2]i^]

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x− ω 2 t−δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x− ω 2 t−δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x− ω 2 t−δ))]i^

Conclusion:

Therefore, the instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

(b)

To determine

The time averaged Poynting vector.

(b)

Expert Solution

Answer to Problem 51P

The time averaged Poynting vector is 12μ0c[E102−E202]i^ .

Explanation of Solution

Calculation:

Consider the relation,

S→=1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t+δ))]i^

The time average of the square of the cosine term is 0.5 so, the time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the time averaged Poynting vector is 12μ0c[E102−E202]i^ .

(c)

To determine

The instantaneous and time average Poynting vector.

(c)

Expert Solution

Answer to Problem 51P

The instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

Explanation of Solution

Given:

The direction of propagation of second wave is E→2=E20cos(k2x+ω2t+δ)j^ .

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x+ ω 2 t+δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x+ ω 2 t+δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x+ ω 2 t+δ))]i^

The time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

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Answer 4

Question

Chapter 30, Problem 51P

(a)

To determine

The instantaneous Poynting vector.

(a)

Expert Solution

Answer to Problem 51P

The instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

Explanation of Solution

Given:

The electric field of first wave is E→1=E10cos(k1x−ω1t)j^ .

The electric field of second wave is E→2=E20cos(k2x−ω2t+δ)j^ .

Formula used:

The expression for instantaneous Poynting vector is given by,

S→=1μ0[E→1+E→2]×1c[[E→1−E→2]i^]

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x− ω 2 t−δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x− ω 2 t−δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x− ω 2 t−δ))]i^

Conclusion:

Therefore, the instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

(b)

To determine

The time averaged Poynting vector.

(b)

Expert Solution

Answer to Problem 51P

The time averaged Poynting vector is 12μ0c[E102−E202]i^ .

Explanation of Solution

Calculation:

Consider the relation,

S→=1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t+δ))]i^

The time average of the square of the cosine term is 0.5 so, the time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the time averaged Poynting vector is 12μ0c[E102−E202]i^ .

(c)

To determine

The instantaneous and time average Poynting vector.

(c)

Expert Solution

Answer to Problem 51P

The instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

Explanation of Solution

Given:

The direction of propagation of second wave is E→2=E20cos(k2x+ω2t+δ)j^ .

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x+ ω 2 t+δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x+ ω 2 t+δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x+ ω 2 t+δ))]i^

The time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

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Answer 5

Chapter 30, Problem 51P

(a)

To determine

The instantaneous Poynting vector.

(a)

Expert Solution

Answer to Problem 51P

The instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

Explanation of Solution

Given:

The electric field of first wave is E→1=E10cos(k1x−ω1t)j^ .

The electric field of second wave is E→2=E20cos(k2x−ω2t+δ)j^ .

Formula used:

The expression for instantaneous Poynting vector is given by,

S→=1μ0[E→1+E→2]×1c[[E→1−E→2]i^]

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x− ω 2 t−δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x− ω 2 t−δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x− ω 2 t−δ))]i^

Conclusion:

Therefore, the instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

(b)

To determine

The time averaged Poynting vector.

(b)

Expert Solution

Answer to Problem 51P

The time averaged Poynting vector is 12μ0c[E102−E202]i^ .

Explanation of Solution

Calculation:

Consider the relation,

S→=1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t+δ))]i^

The time average of the square of the cosine term is 0.5 so, the time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the time averaged Poynting vector is 12μ0c[E102−E202]i^ .

(c)

To determine

The instantaneous and time average Poynting vector.

(c)

Expert Solution

Answer to Problem 51P

The instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

Explanation of Solution

Given:

The direction of propagation of second wave is E→2=E20cos(k2x+ω2t+δ)j^ .

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x+ ω 2 t+δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x+ ω 2 t+δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x+ ω 2 t+δ))]i^

The time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

Answer 6

Chapter 30, Problem 51P

(a)

To determine

The instantaneous Poynting vector.

(a)

Expert Solution

Answer to Problem 51P

The instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

Explanation of Solution

Given:

The electric field of first wave is E→1=E10cos(k1x−ω1t)j^ .

The electric field of second wave is E→2=E20cos(k2x−ω2t+δ)j^ .

Formula used:

The expression for instantaneous Poynting vector is given by,

S→=1μ0[E→1+E→2]×1c[[E→1−E→2]i^]

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x− ω 2 t−δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x− ω 2 t−δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x− ω 2 t−δ))]i^

Conclusion:

Therefore, the instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

Answer 7

Chapter 30, Problem 51P

(a)

To determine

The instantaneous Poynting vector.

Answer 8

(a)

Expert Solution

Answer to Problem 51P

The instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

The electric field of first wave is E→1=E10cos(k1x−ω1t)j^ .

The electric field of second wave is E→2=E20cos(k2x−ω2t+δ)j^ .

Formula used:

The expression for instantaneous Poynting vector is given by,

S→=1μ0[E→1+E→2]×1c[[E→1−E→2]i^]

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x− ω 2 t−δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x− ω 2 t−δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x− ω 2 t−δ))]i^

Conclusion:

Therefore, the instantaneous Poynting vector is 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t−δ))]i^ .

Answer 13

(b)

To determine

The time averaged Poynting vector.

(b)

Expert Solution

Answer to Problem 51P

The time averaged Poynting vector is 12μ0c[E102−E202]i^ .

Explanation of Solution

Calculation:

Consider the relation,

S→=1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t+δ))]i^

The time average of the square of the cosine term is 0.5 so, the time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the time averaged Poynting vector is 12μ0c[E102−E202]i^ .

Answer 14

(b)

To determine

The time averaged Poynting vector.

Answer 15

(b)

Expert Solution

Answer to Problem 51P

The time averaged Poynting vector is 12μ0c[E102−E202]i^ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

Consider the relation,

S→=1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x− ω 2t+δ))]i^

The time average of the square of the cosine term is 0.5 so, the time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the time averaged Poynting vector is 12μ0c[E102−E202]i^ .

Answer 20

(c)

To determine

The instantaneous and time average Poynting vector.

(c)

Expert Solution

Answer to Problem 51P

The instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

Explanation of Solution

Given:

The direction of propagation of second wave is E→2=E20cos(k2x+ω2t+δ)j^ .

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x+ ω 2 t+δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x+ ω 2 t+δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x+ ω 2 t+δ))]i^

The time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

Answer 21

(c)

To determine

The instantaneous and time average Poynting vector.

Answer 22

(c)

Expert Solution

Answer to Problem 51P

The instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given:

The direction of propagation of second wave is E→2=E20cos(k2x+ω2t+δ)j^ .

Calculation:

The instantaneous Poynting vector is calculated as,

S→=1μ0[ E →1+ E →2]×1c[[ E → 1− E → 2]i^]=1μ0[( E 10cos( k 1 x− ω 1 t) j ^)+( E 20cos( k 2 x+ ω 2 t+δ) j ^)]×1c[( E 10cos( k 1 x− ω 1 t) j ^)−( E 20cos( k 2 x+ ω 2 t+δ) j ^)]i^=1μ0c[E102(cos( k 1 x− ω 1 t))−E202(cos( k 2 x+ ω 2 t+δ))]i^

The time averaged Poynting vector is,

S→av=12μ0c[E102−E202]i^

Conclusion:

Therefore, the instantaneous and time averaged Poynting vector are 1μ0c[E102(cos( k 1x− ω 1t))−E202(cos( k 2x+ ω 2t+δ))]i^ and 12μ0c[E102−E202]i^ respectively.