To prove □ ⋅ F = − μ 0 J .

Question

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Answer 1

Question

Chapter 15, Problem 15.111P

To determine

To prove □⋅F=−μ0J.

Expert Solution & Answer

Answer to Problem 15.111P

It is proven that □⋅F=−μ0J.

Explanation of Solution

Write the expression for Maxwell’s equation

∇×B−1c2∂E∂t=μ0J (I)

The above equation can be expressed as

|ijk∂∂x1∂∂x2∂∂x3B1B2B3|−1c∂∂(ct)(E1i+E2j+E3k)=μ0(J1i+J2j+J3k)

Expand the above equation.

[∂B3∂x2−∂B2∂x3−1c∂E1∂x4]i+[∂B1∂x3−∂B3∂x1−1c∂E2∂x4]j+[∂B2∂x1−∂B1∂x2−1c∂E3∂x4]k=(μ0J1)i+(μ0J2)j+(μ0J3)k (II)

Equate the i^ components of the above equation.

0+∂B3∂x2−∂B2∂x3−∂∂x4(E1c)=μ0J1 (III)

Equate the j^ components of equation (II).

−∂B3∂x1+0+∂B1∂x3−∂∂x4(E2c)=μ0J2 (IV)

Equate the k^ components of equation (II).

∂B2∂x1−∂B1∂x2+0−∂∂x4(E3c)=μ0J3 (V)

Write the expression for Maxwell’s equation.

∇⋅E=1ε0k

The above equation can be written as

∂E1∂x1+∂E2∂x2+∂E3∂x3=1ε0k

Since J4=ck, substitute k=J4c in the above equation.

∂E1∂x1+∂E2∂x2+∂E3∂x3=1ε0J4c1c∂E1∂x1+1c∂E2∂x2+1c∂E3∂x3=1ε0J4c2

Use c2=1μ0ε0 in the above equation.

∂∂x1(E1c)+∂∂x2(E2c)+∂∂x3(E3c)=1ε0(μ0ε0)J4∂∂x1(E1c)+∂∂x2(E2c)+∂∂x3(E3c)=μ0J4 (VI)

Write the matrix form of F.

F=[0B3−B2−E1c−B30B1−E2cB2−B10−E3cE1cE2cE3c0] (VII)

Write the matrix for G.

G=[+10000+10000+10000−1] (VIII)

Write the equation for □GF.

□GF=(∂∂x1∂∂x2∂∂x3∂∂x3−∂∂x4)[0B3−B2−E1c−B30B1−E2cB2−B10−E3c−E1c−E2c−E3c0]=(0+∂∂x2(−B3)+∂∂x3(B2)−∂∂x4(−E1c)∂∂x1(B3)+0+∂∂x4(−E2c)∂∂x1(−B2)+∂∂x2(B1)+0−∂∂x4(E3c)∂∂x1(−E1c)+∂∂x2(−E2c)+∂∂x3(−E3c)+0)

Since □⋅F=□GF, the matrix form of □GF is given by

□GF=[0+∂∂x2(−B3)+∂∂x3(B2)+∂∂x4(E1c)∂∂x1(B3)+0+∂∂x3(−B1)+∂∂x4(E2c)∂∂x1(−B2)+∂∂x2(B1)+0+∂∂x4(E3c)∂∂x1(−E1c)+∂∂x2(−E2c)+∂∂x3(−E3c)+0] (IX)

Write the matrix form of GF.

GF=[+10000+10000+10000−1][0B3−B2−E1c−B30B1−E2cB2−B10−E3cE1cE2cE3c0]=[0+0+0+0B3+0+0+0−B2+0+0+0−E1c+0+0+00−B3+0+00+0+0+00+B1+0+00−E2c+0+00+0+B2+00+0−B1+00+0+0+00+0−E3c+00+0+0−E1c0+0+0−E2c0+0+0−E3c0+0+0+0]=[0B3−B2−E1c−B30B1−E2cB2−B10−E3c−E1c−E2c−E3c0] (X)

Since □=(∂∂x1∂∂x2∂∂x3−∂∂x4) equation (IX) is written as

□⋅F=[−(0+∂∂x2(B2)+∂∂x3(−B2)+∂∂x4(−E1c))−(∂∂x1(−B3)+0+∂∂x3(B1)+∂∂x4(−E2c))−(∂∂x1(B2)+∂∂x2(−B1)+0+∂∂x4(−E3c)−(∂∂x1(E1c)+∂∂x2(E2c)+∂∂x3(E3c)+0]=[−μ0J1−μ0J2−μ0J3−μ0J4]=−μ0[J1J2J3J4] (XI)

Since [J1J2J3J4]=J, equation (XI) is written as

□⋅F=μ0J

Conclusion:

Therefore, it is proven that □⋅F=−μ0J.

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Answer 2

Question

Chapter 15, Problem 15.111P

To determine

To prove □⋅F=−μ0J.

Expert Solution & Answer

Answer to Problem 15.111P

It is proven that □⋅F=−μ0J.

Explanation of Solution

Write the expression for Maxwell’s equation

∇×B−1c2∂E∂t=μ0J (I)

The above equation can be expressed as

|ijk∂∂x1∂∂x2∂∂x3B1B2B3|−1c∂∂(ct)(E1i+E2j+E3k)=μ0(J1i+J2j+J3k)

Expand the above equation.

[∂B3∂x2−∂B2∂x3−1c∂E1∂x4]i+[∂B1∂x3−∂B3∂x1−1c∂E2∂x4]j+[∂B2∂x1−∂B1∂x2−1c∂E3∂x4]k=(μ0J1)i+(μ0J2)j+(μ0J3)k (II)

Equate the i^ components of the above equation.

0+∂B3∂x2−∂B2∂x3−∂∂x4(E1c)=μ0J1 (III)

Equate the j^ components of equation (II).

−∂B3∂x1+0+∂B1∂x3−∂∂x4(E2c)=μ0J2 (IV)

Equate the k^ components of equation (II).

∂B2∂x1−∂B1∂x2+0−∂∂x4(E3c)=μ0J3 (V)

Write the expression for Maxwell’s equation.

∇⋅E=1ε0k

The above equation can be written as

∂E1∂x1+∂E2∂x2+∂E3∂x3=1ε0k

Since J4=ck, substitute k=J4c in the above equation.

∂E1∂x1+∂E2∂x2+∂E3∂x3=1ε0J4c1c∂E1∂x1+1c∂E2∂x2+1c∂E3∂x3=1ε0J4c2

Use c2=1μ0ε0 in the above equation.

∂∂x1(E1c)+∂∂x2(E2c)+∂∂x3(E3c)=1ε0(μ0ε0)J4∂∂x1(E1c)+∂∂x2(E2c)+∂∂x3(E3c)=μ0J4 (VI)

Write the matrix form of F.

F=[0B3−B2−E1c−B30B1−E2cB2−B10−E3cE1cE2cE3c0] (VII)

Write the matrix for G.

G=[+10000+10000+10000−1] (VIII)

Write the equation for □GF.

□GF=(∂∂x1∂∂x2∂∂x3∂∂x3−∂∂x4)[0B3−B2−E1c−B30B1−E2cB2−B10−E3c−E1c−E2c−E3c0]=(0+∂∂x2(−B3)+∂∂x3(B2)−∂∂x4(−E1c)∂∂x1(B3)+0+∂∂x4(−E2c)∂∂x1(−B2)+∂∂x2(B1)+0−∂∂x4(E3c)∂∂x1(−E1c)+∂∂x2(−E2c)+∂∂x3(−E3c)+0)

Since □⋅F=□GF, the matrix form of □GF is given by

□GF=[0+∂∂x2(−B3)+∂∂x3(B2)+∂∂x4(E1c)∂∂x1(B3)+0+∂∂x3(−B1)+∂∂x4(E2c)∂∂x1(−B2)+∂∂x2(B1)+0+∂∂x4(E3c)∂∂x1(−E1c)+∂∂x2(−E2c)+∂∂x3(−E3c)+0] (IX)

Write the matrix form of GF.

GF=[+10000+10000+10000−1][0B3−B2−E1c−B30B1−E2cB2−B10−E3cE1cE2cE3c0]=[0+0+0+0B3+0+0+0−B2+0+0+0−E1c+0+0+00−B3+0+00+0+0+00+B1+0+00−E2c+0+00+0+B2+00+0−B1+00+0+0+00+0−E3c+00+0+0−E1c0+0+0−E2c0+0+0−E3c0+0+0+0]=[0B3−B2−E1c−B30B1−E2cB2−B10−E3c−E1c−E2c−E3c0] (X)

Since □=(∂∂x1∂∂x2∂∂x3−∂∂x4) equation (IX) is written as

□⋅F=[−(0+∂∂x2(B2)+∂∂x3(−B2)+∂∂x4(−E1c))−(∂∂x1(−B3)+0+∂∂x3(B1)+∂∂x4(−E2c))−(∂∂x1(B2)+∂∂x2(−B1)+0+∂∂x4(−E3c)−(∂∂x1(E1c)+∂∂x2(E2c)+∂∂x3(E3c)+0]=[−μ0J1−μ0J2−μ0J3−μ0J4]=−μ0[J1J2J3J4] (XI)

Since [J1J2J3J4]=J, equation (XI) is written as

□⋅F=μ0J

Conclusion:

Therefore, it is proven that □⋅F=−μ0J.

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Answer 3

Question

Chapter 15, Problem 15.111P

To determine

To prove □⋅F=−μ0J.

Expert Solution & Answer

Answer to Problem 15.111P

It is proven that □⋅F=−μ0J.

Explanation of Solution

Write the expression for Maxwell’s equation

∇×B−1c2∂E∂t=μ0J (I)

The above equation can be expressed as

|ijk∂∂x1∂∂x2∂∂x3B1B2B3|−1c∂∂(ct)(E1i+E2j+E3k)=μ0(J1i+J2j+J3k)

Expand the above equation.

[∂B3∂x2−∂B2∂x3−1c∂E1∂x4]i+[∂B1∂x3−∂B3∂x1−1c∂E2∂x4]j+[∂B2∂x1−∂B1∂x2−1c∂E3∂x4]k=(μ0J1)i+(μ0J2)j+(μ0J3)k (II)

Equate the i^ components of the above equation.

0+∂B3∂x2−∂B2∂x3−∂∂x4(E1c)=μ0J1 (III)

Equate the j^ components of equation (II).

−∂B3∂x1+0+∂B1∂x3−∂∂x4(E2c)=μ0J2 (IV)

Equate the k^ components of equation (II).

∂B2∂x1−∂B1∂x2+0−∂∂x4(E3c)=μ0J3 (V)

Write the expression for Maxwell’s equation.

∇⋅E=1ε0k

The above equation can be written as

∂E1∂x1+∂E2∂x2+∂E3∂x3=1ε0k

Since J4=ck, substitute k=J4c in the above equation.

∂E1∂x1+∂E2∂x2+∂E3∂x3=1ε0J4c1c∂E1∂x1+1c∂E2∂x2+1c∂E3∂x3=1ε0J4c2

Use c2=1μ0ε0 in the above equation.

∂∂x1(E1c)+∂∂x2(E2c)+∂∂x3(E3c)=1ε0(μ0ε0)J4∂∂x1(E1c)+∂∂x2(E2c)+∂∂x3(E3c)=μ0J4 (VI)

Write the matrix form of F.

F=[0B3−B2−E1c−B30B1−E2cB2−B10−E3cE1cE2cE3c0] (VII)

Write the matrix for G.

G=[+10000+10000+10000−1] (VIII)

Write the equation for □GF.

□GF=(∂∂x1∂∂x2∂∂x3∂∂x3−∂∂x4)[0B3−B2−E1c−B30B1−E2cB2−B10−E3c−E1c−E2c−E3c0]=(0+∂∂x2(−B3)+∂∂x3(B2)−∂∂x4(−E1c)∂∂x1(B3)+0+∂∂x4(−E2c)∂∂x1(−B2)+∂∂x2(B1)+0−∂∂x4(E3c)∂∂x1(−E1c)+∂∂x2(−E2c)+∂∂x3(−E3c)+0)

Since □⋅F=□GF, the matrix form of □GF is given by

□GF=[0+∂∂x2(−B3)+∂∂x3(B2)+∂∂x4(E1c)∂∂x1(B3)+0+∂∂x3(−B1)+∂∂x4(E2c)∂∂x1(−B2)+∂∂x2(B1)+0+∂∂x4(E3c)∂∂x1(−E1c)+∂∂x2(−E2c)+∂∂x3(−E3c)+0] (IX)

Write the matrix form of GF.

GF=[+10000+10000+10000−1][0B3−B2−E1c−B30B1−E2cB2−B10−E3cE1cE2cE3c0]=[0+0+0+0B3+0+0+0−B2+0+0+0−E1c+0+0+00−B3+0+00+0+0+00+B1+0+00−E2c+0+00+0+B2+00+0−B1+00+0+0+00+0−E3c+00+0+0−E1c0+0+0−E2c0+0+0−E3c0+0+0+0]=[0B3−B2−E1c−B30B1−E2cB2−B10−E3c−E1c−E2c−E3c0] (X)

Since □=(∂∂x1∂∂x2∂∂x3−∂∂x4) equation (IX) is written as

□⋅F=[−(0+∂∂x2(B2)+∂∂x3(−B2)+∂∂x4(−E1c))−(∂∂x1(−B3)+0+∂∂x3(B1)+∂∂x4(−E2c))−(∂∂x1(B2)+∂∂x2(−B1)+0+∂∂x4(−E3c)−(∂∂x1(E1c)+∂∂x2(E2c)+∂∂x3(E3c)+0]=[−μ0J1−μ0J2−μ0J3−μ0J4]=−μ0[J1J2J3J4] (XI)

Since [J1J2J3J4]=J, equation (XI) is written as

□⋅F=μ0J

Conclusion:

Therefore, it is proven that □⋅F=−μ0J.

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Answer 4

Question

Chapter 15, Problem 15.111P

To determine

To prove □⋅F=−μ0J.

Expert Solution & Answer

Answer to Problem 15.111P

It is proven that □⋅F=−μ0J.

Explanation of Solution

Write the expression for Maxwell’s equation

∇×B−1c2∂E∂t=μ0J (I)

The above equation can be expressed as

|ijk∂∂x1∂∂x2∂∂x3B1B2B3|−1c∂∂(ct)(E1i+E2j+E3k)=μ0(J1i+J2j+J3k)

Expand the above equation.

[∂B3∂x2−∂B2∂x3−1c∂E1∂x4]i+[∂B1∂x3−∂B3∂x1−1c∂E2∂x4]j+[∂B2∂x1−∂B1∂x2−1c∂E3∂x4]k=(μ0J1)i+(μ0J2)j+(μ0J3)k (II)

Equate the i^ components of the above equation.

0+∂B3∂x2−∂B2∂x3−∂∂x4(E1c)=μ0J1 (III)

Equate the j^ components of equation (II).

−∂B3∂x1+0+∂B1∂x3−∂∂x4(E2c)=μ0J2 (IV)

Equate the k^ components of equation (II).

∂B2∂x1−∂B1∂x2+0−∂∂x4(E3c)=μ0J3 (V)

Write the expression for Maxwell’s equation.

∇⋅E=1ε0k

The above equation can be written as

∂E1∂x1+∂E2∂x2+∂E3∂x3=1ε0k

Since J4=ck, substitute k=J4c in the above equation.

∂E1∂x1+∂E2∂x2+∂E3∂x3=1ε0J4c1c∂E1∂x1+1c∂E2∂x2+1c∂E3∂x3=1ε0J4c2

Use c2=1μ0ε0 in the above equation.

∂∂x1(E1c)+∂∂x2(E2c)+∂∂x3(E3c)=1ε0(μ0ε0)J4∂∂x1(E1c)+∂∂x2(E2c)+∂∂x3(E3c)=μ0J4 (VI)

Write the matrix form of F.

F=[0B3−B2−E1c−B30B1−E2cB2−B10−E3cE1cE2cE3c0] (VII)

Write the matrix for G.

G=[+10000+10000+10000−1] (VIII)

Write the equation for □GF.

□GF=(∂∂x1∂∂x2∂∂x3∂∂x3−∂∂x4)[0B3−B2−E1c−B30B1−E2cB2−B10−E3c−E1c−E2c−E3c0]=(0+∂∂x2(−B3)+∂∂x3(B2)−∂∂x4(−E1c)∂∂x1(B3)+0+∂∂x4(−E2c)∂∂x1(−B2)+∂∂x2(B1)+0−∂∂x4(E3c)∂∂x1(−E1c)+∂∂x2(−E2c)+∂∂x3(−E3c)+0)

Since □⋅F=□GF, the matrix form of □GF is given by

□GF=[0+∂∂x2(−B3)+∂∂x3(B2)+∂∂x4(E1c)∂∂x1(B3)+0+∂∂x3(−B1)+∂∂x4(E2c)∂∂x1(−B2)+∂∂x2(B1)+0+∂∂x4(E3c)∂∂x1(−E1c)+∂∂x2(−E2c)+∂∂x3(−E3c)+0] (IX)

Write the matrix form of GF.

GF=[+10000+10000+10000−1][0B3−B2−E1c−B30B1−E2cB2−B10−E3cE1cE2cE3c0]=[0+0+0+0B3+0+0+0−B2+0+0+0−E1c+0+0+00−B3+0+00+0+0+00+B1+0+00−E2c+0+00+0+B2+00+0−B1+00+0+0+00+0−E3c+00+0+0−E1c0+0+0−E2c0+0+0−E3c0+0+0+0]=[0B3−B2−E1c−B30B1−E2cB2−B10−E3c−E1c−E2c−E3c0] (X)

Since □=(∂∂x1∂∂x2∂∂x3−∂∂x4) equation (IX) is written as

□⋅F=[−(0+∂∂x2(B2)+∂∂x3(−B2)+∂∂x4(−E1c))−(∂∂x1(−B3)+0+∂∂x3(B1)+∂∂x4(−E2c))−(∂∂x1(B2)+∂∂x2(−B1)+0+∂∂x4(−E3c)−(∂∂x1(E1c)+∂∂x2(E2c)+∂∂x3(E3c)+0]=[−μ0J1−μ0J2−μ0J3−μ0J4]=−μ0[J1J2J3J4] (XI)

Since [J1J2J3J4]=J, equation (XI) is written as

□⋅F=μ0J

Conclusion:

Therefore, it is proven that □⋅F=−μ0J.

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Answer 5

Chapter 15, Problem 15.111P

To determine

To prove □⋅F=−μ0J.

Expert Solution & Answer

Answer to Problem 15.111P

It is proven that □⋅F=−μ0J.

Explanation of Solution

Write the expression for Maxwell’s equation

∇×B−1c2∂E∂t=μ0J (I)

The above equation can be expressed as

|ijk∂∂x1∂∂x2∂∂x3B1B2B3|−1c∂∂(ct)(E1i+E2j+E3k)=μ0(J1i+J2j+J3k)

Expand the above equation.

[∂B3∂x2−∂B2∂x3−1c∂E1∂x4]i+[∂B1∂x3−∂B3∂x1−1c∂E2∂x4]j+[∂B2∂x1−∂B1∂x2−1c∂E3∂x4]k=(μ0J1)i+(μ0J2)j+(μ0J3)k (II)

Equate the i^ components of the above equation.

0+∂B3∂x2−∂B2∂x3−∂∂x4(E1c)=μ0J1 (III)

Equate the j^ components of equation (II).

−∂B3∂x1+0+∂B1∂x3−∂∂x4(E2c)=μ0J2 (IV)

Equate the k^ components of equation (II).

∂B2∂x1−∂B1∂x2+0−∂∂x4(E3c)=μ0J3 (V)

Write the expression for Maxwell’s equation.

∇⋅E=1ε0k

The above equation can be written as

∂E1∂x1+∂E2∂x2+∂E3∂x3=1ε0k

Since J4=ck, substitute k=J4c in the above equation.

∂E1∂x1+∂E2∂x2+∂E3∂x3=1ε0J4c1c∂E1∂x1+1c∂E2∂x2+1c∂E3∂x3=1ε0J4c2

Use c2=1μ0ε0 in the above equation.

∂∂x1(E1c)+∂∂x2(E2c)+∂∂x3(E3c)=1ε0(μ0ε0)J4∂∂x1(E1c)+∂∂x2(E2c)+∂∂x3(E3c)=μ0J4 (VI)

Write the matrix form of F.

F=[0B3−B2−E1c−B30B1−E2cB2−B10−E3cE1cE2cE3c0] (VII)

Write the matrix for G.

G=[+10000+10000+10000−1] (VIII)

Write the equation for □GF.

□GF=(∂∂x1∂∂x2∂∂x3∂∂x3−∂∂x4)[0B3−B2−E1c−B30B1−E2cB2−B10−E3c−E1c−E2c−E3c0]=(0+∂∂x2(−B3)+∂∂x3(B2)−∂∂x4(−E1c)∂∂x1(B3)+0+∂∂x4(−E2c)∂∂x1(−B2)+∂∂x2(B1)+0−∂∂x4(E3c)∂∂x1(−E1c)+∂∂x2(−E2c)+∂∂x3(−E3c)+0)

Since □⋅F=□GF, the matrix form of □GF is given by

□GF=[0+∂∂x2(−B3)+∂∂x3(B2)+∂∂x4(E1c)∂∂x1(B3)+0+∂∂x3(−B1)+∂∂x4(E2c)∂∂x1(−B2)+∂∂x2(B1)+0+∂∂x4(E3c)∂∂x1(−E1c)+∂∂x2(−E2c)+∂∂x3(−E3c)+0] (IX)

Write the matrix form of GF.

GF=[+10000+10000+10000−1][0B3−B2−E1c−B30B1−E2cB2−B10−E3cE1cE2cE3c0]=[0+0+0+0B3+0+0+0−B2+0+0+0−E1c+0+0+00−B3+0+00+0+0+00+B1+0+00−E2c+0+00+0+B2+00+0−B1+00+0+0+00+0−E3c+00+0+0−E1c0+0+0−E2c0+0+0−E3c0+0+0+0]=[0B3−B2−E1c−B30B1−E2cB2−B10−E3c−E1c−E2c−E3c0] (X)

Since □=(∂∂x1∂∂x2∂∂x3−∂∂x4) equation (IX) is written as

□⋅F=[−(0+∂∂x2(B2)+∂∂x3(−B2)+∂∂x4(−E1c))−(∂∂x1(−B3)+0+∂∂x3(B1)+∂∂x4(−E2c))−(∂∂x1(B2)+∂∂x2(−B1)+0+∂∂x4(−E3c)−(∂∂x1(E1c)+∂∂x2(E2c)+∂∂x3(E3c)+0]=[−μ0J1−μ0J2−μ0J3−μ0J4]=−μ0[J1J2J3J4] (XI)

Since [J1J2J3J4]=J, equation (XI) is written as

□⋅F=μ0J

Conclusion:

Therefore, it is proven that □⋅F=−μ0J.

Answer 6

Chapter 15, Problem 15.111P

To determine

To prove □⋅F=−μ0J.

Expert Solution & Answer

Answer to Problem 15.111P

It is proven that □⋅F=−μ0J.

Explanation of Solution

Write the expression for Maxwell’s equation

∇×B−1c2∂E∂t=μ0J (I)

The above equation can be expressed as

|ijk∂∂x1∂∂x2∂∂x3B1B2B3|−1c∂∂(ct)(E1i+E2j+E3k)=μ0(J1i+J2j+J3k)

Expand the above equation.

[∂B3∂x2−∂B2∂x3−1c∂E1∂x4]i+[∂B1∂x3−∂B3∂x1−1c∂E2∂x4]j+[∂B2∂x1−∂B1∂x2−1c∂E3∂x4]k=(μ0J1)i+(μ0J2)j+(μ0J3)k (II)

Equate the i^ components of the above equation.

0+∂B3∂x2−∂B2∂x3−∂∂x4(E1c)=μ0J1 (III)

Equate the j^ components of equation (II).

−∂B3∂x1+0+∂B1∂x3−∂∂x4(E2c)=μ0J2 (IV)

Equate the k^ components of equation (II).

∂B2∂x1−∂B1∂x2+0−∂∂x4(E3c)=μ0J3 (V)

Write the expression for Maxwell’s equation.

∇⋅E=1ε0k

The above equation can be written as

∂E1∂x1+∂E2∂x2+∂E3∂x3=1ε0k

Since J4=ck, substitute k=J4c in the above equation.

∂E1∂x1+∂E2∂x2+∂E3∂x3=1ε0J4c1c∂E1∂x1+1c∂E2∂x2+1c∂E3∂x3=1ε0J4c2

Use c2=1μ0ε0 in the above equation.

∂∂x1(E1c)+∂∂x2(E2c)+∂∂x3(E3c)=1ε0(μ0ε0)J4∂∂x1(E1c)+∂∂x2(E2c)+∂∂x3(E3c)=μ0J4 (VI)

Write the matrix form of F.

F=[0B3−B2−E1c−B30B1−E2cB2−B10−E3cE1cE2cE3c0] (VII)

Write the matrix for G.

G=[+10000+10000+10000−1] (VIII)

Write the equation for □GF.

□GF=(∂∂x1∂∂x2∂∂x3∂∂x3−∂∂x4)[0B3−B2−E1c−B30B1−E2cB2−B10−E3c−E1c−E2c−E3c0]=(0+∂∂x2(−B3)+∂∂x3(B2)−∂∂x4(−E1c)∂∂x1(B3)+0+∂∂x4(−E2c)∂∂x1(−B2)+∂∂x2(B1)+0−∂∂x4(E3c)∂∂x1(−E1c)+∂∂x2(−E2c)+∂∂x3(−E3c)+0)

Since □⋅F=□GF, the matrix form of □GF is given by

□GF=[0+∂∂x2(−B3)+∂∂x3(B2)+∂∂x4(E1c)∂∂x1(B3)+0+∂∂x3(−B1)+∂∂x4(E2c)∂∂x1(−B2)+∂∂x2(B1)+0+∂∂x4(E3c)∂∂x1(−E1c)+∂∂x2(−E2c)+∂∂x3(−E3c)+0] (IX)

Write the matrix form of GF.

GF=[+10000+10000+10000−1][0B3−B2−E1c−B30B1−E2cB2−B10−E3cE1cE2cE3c0]=[0+0+0+0B3+0+0+0−B2+0+0+0−E1c+0+0+00−B3+0+00+0+0+00+B1+0+00−E2c+0+00+0+B2+00+0−B1+00+0+0+00+0−E3c+00+0+0−E1c0+0+0−E2c0+0+0−E3c0+0+0+0]=[0B3−B2−E1c−B30B1−E2cB2−B10−E3c−E1c−E2c−E3c0] (X)

Since □=(∂∂x1∂∂x2∂∂x3−∂∂x4) equation (IX) is written as

□⋅F=[−(0+∂∂x2(B2)+∂∂x3(−B2)+∂∂x4(−E1c))−(∂∂x1(−B3)+0+∂∂x3(B1)+∂∂x4(−E2c))−(∂∂x1(B2)+∂∂x2(−B1)+0+∂∂x4(−E3c)−(∂∂x1(E1c)+∂∂x2(E2c)+∂∂x3(E3c)+0]=[−μ0J1−μ0J2−μ0J3−μ0J4]=−μ0[J1J2J3J4] (XI)

Since [J1J2J3J4]=J, equation (XI) is written as

□⋅F=μ0J

Conclusion:

Therefore, it is proven that □⋅F=−μ0J.

Answer 7

Chapter 15, Problem 15.111P

To determine

To prove □⋅F=−μ0J.

Answer 8

Expert Solution & Answer

Answer to Problem 15.111P

It is proven that □⋅F=−μ0J.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Write the expression for Maxwell’s equation

∇×B−1c2∂E∂t=μ0J (I)

The above equation can be expressed as

|ijk∂∂x1∂∂x2∂∂x3B1B2B3|−1c∂∂(ct)(E1i+E2j+E3k)=μ0(J1i+J2j+J3k)

Expand the above equation.

[∂B3∂x2−∂B2∂x3−1c∂E1∂x4]i+[∂B1∂x3−∂B3∂x1−1c∂E2∂x4]j+[∂B2∂x1−∂B1∂x2−1c∂E3∂x4]k=(μ0J1)i+(μ0J2)j+(μ0J3)k (II)

Equate the i^ components of the above equation.

0+∂B3∂x2−∂B2∂x3−∂∂x4(E1c)=μ0J1 (III)

Equate the j^ components of equation (II).

−∂B3∂x1+0+∂B1∂x3−∂∂x4(E2c)=μ0J2 (IV)

Equate the k^ components of equation (II).

∂B2∂x1−∂B1∂x2+0−∂∂x4(E3c)=μ0J3 (V)

Write the expression for Maxwell’s equation.

∇⋅E=1ε0k

The above equation can be written as

∂E1∂x1+∂E2∂x2+∂E3∂x3=1ε0k

Since J4=ck, substitute k=J4c in the above equation.

∂E1∂x1+∂E2∂x2+∂E3∂x3=1ε0J4c1c∂E1∂x1+1c∂E2∂x2+1c∂E3∂x3=1ε0J4c2

Use c2=1μ0ε0 in the above equation.

∂∂x1(E1c)+∂∂x2(E2c)+∂∂x3(E3c)=1ε0(μ0ε0)J4∂∂x1(E1c)+∂∂x2(E2c)+∂∂x3(E3c)=μ0J4 (VI)

Write the matrix form of F.

F=[0B3−B2−E1c−B30B1−E2cB2−B10−E3cE1cE2cE3c0] (VII)

Write the matrix for G.

G=[+10000+10000+10000−1] (VIII)

Write the equation for □GF.

□GF=(∂∂x1∂∂x2∂∂x3∂∂x3−∂∂x4)[0B3−B2−E1c−B30B1−E2cB2−B10−E3c−E1c−E2c−E3c0]=(0+∂∂x2(−B3)+∂∂x3(B2)−∂∂x4(−E1c)∂∂x1(B3)+0+∂∂x4(−E2c)∂∂x1(−B2)+∂∂x2(B1)+0−∂∂x4(E3c)∂∂x1(−E1c)+∂∂x2(−E2c)+∂∂x3(−E3c)+0)

Since □⋅F=□GF, the matrix form of □GF is given by

□GF=[0+∂∂x2(−B3)+∂∂x3(B2)+∂∂x4(E1c)∂∂x1(B3)+0+∂∂x3(−B1)+∂∂x4(E2c)∂∂x1(−B2)+∂∂x2(B1)+0+∂∂x4(E3c)∂∂x1(−E1c)+∂∂x2(−E2c)+∂∂x3(−E3c)+0] (IX)

Write the matrix form of GF.

GF=[+10000+10000+10000−1][0B3−B2−E1c−B30B1−E2cB2−B10−E3cE1cE2cE3c0]=[0+0+0+0B3+0+0+0−B2+0+0+0−E1c+0+0+00−B3+0+00+0+0+00+B1+0+00−E2c+0+00+0+B2+00+0−B1+00+0+0+00+0−E3c+00+0+0−E1c0+0+0−E2c0+0+0−E3c0+0+0+0]=[0B3−B2−E1c−B30B1−E2cB2−B10−E3c−E1c−E2c−E3c0] (X)

Since □=(∂∂x1∂∂x2∂∂x3−∂∂x4) equation (IX) is written as

□⋅F=[−(0+∂∂x2(B2)+∂∂x3(−B2)+∂∂x4(−E1c))−(∂∂x1(−B3)+0+∂∂x3(B1)+∂∂x4(−E2c))−(∂∂x1(B2)+∂∂x2(−B1)+0+∂∂x4(−E3c)−(∂∂x1(E1c)+∂∂x2(E2c)+∂∂x3(E3c)+0]=[−μ0J1−μ0J2−μ0J3−μ0J4]=−μ0[J1J2J3J4] (XI)