The magnetic flux through the top face of the cube

Question

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Answer 1

Question

Chapter 31, Problem 64PQ

(a)

To determine

The magnetic flux through the top face of the cube

(a)

Expert Solution

Answer to Problem 64PQ

The magnetic flux through the top face of the cube is −5.00×10−2 Wb in downward direction.

Explanation of Solution

Write the expression of the magnetic flux.

ϕ=B→⋅A→ (I)

Here, ϕ is the magnetic flux, B→ is the magnetic field vector and A→ is the normal area vector.

Write the expression for the normal area vector of the top face of the cube.

A→=d2k^ (II)

Here, A→ is the normal vector of the top face of the cube.

Conclusion:

Substitute 10.0cm for d in equation (II).

A→=(10cm)2k^=100k^ cm2

Substitute 100.0 cm2k^ for A→ and (3i^−5j^−2k^)T for B→ in equation (I).

ϕ=((3i^−5j^−2k^)T)⋅(100 cm2)j^=((3i^−5j^−2k^)T)⋅(100 cm2(1 m2104 cm2))j^=−5.00×10−2 Wb

Thus, the magnetic flux through the top face of the cube is −5.00×10−2 Wb in downward direction.

(b)

To determine

The total magnetic flux through all six faces of a cube.

(b)

Expert Solution

Answer to Problem 64PQ

The net magnetic flux through all the six faces of the cube is 0 Wb.

Explanation of Solution

Gauss’s law for magnetic states that the total magentic flux through a closed surface is equal to zero, i.e, all the line coming outside of the cube will ultimately go inside the body, becouse they from closed loops. Mathemeatically, it can be shown that the total magnetic flux is zero.

Write the expression for the net magnetic flux

ϕnet=ϕ1+ϕ2+ϕ3+ϕ4+ϕ5+ϕ6 (III)

Here, ϕnet is the net magnetic flux, ϕ1 , ϕ2 , ϕ3 , ϕ4 , ϕ5 and ϕ6 are the flux through surface 1,2,3,4,5 and 6 respectively.

Substitute ϕ1 for ϕ, (3i^−5j^−2k^) T for B→ and 100 cm2 i^ for A→ in equation (I).

ϕ1=(3i^−5j^−2k^) T⋅(100 cm2i^)=(3i^−5j^−2k^) T⋅(100 cm2(1 m2104 cm2))i^=0.03 Wb

Substitute ϕ2 for ϕ, (3i^−5j^−2k^) T for B→ and −100 cm2 i^ for A→ in equation (I).

ϕ2=(3i^−5j^−2k^) T⋅(−100 cm2i^)=(3i^−5j^−2k^) T⋅(−100 cm2(1 m2104 cm2))i^=−0.03 Wb

Substitute ϕ3 for ϕ, (3i^−5j^−2k^) T for B→ and 100 cm2 j^ for A→ in equation (I).

ϕ3=(3i^−5j^−2k^) T⋅(100 cm2j^)=(3i^−5j^−2k^) T⋅(100 cm2(1 m2104 cm2))j^=−0.05 Wb

Substitute ϕ4 for ϕ, (3i^−5j^−2k^) T for B→ and −100 cm2 j^ for A→ in equation (I).

ϕ4=(3i^−5j^−2k^) T⋅(−100 cm2j^)=(3i^−5j^−2k^) T⋅(−100 cm2(1 m2104 cm2))j^=0.05 Wb

Substitute ϕ5 for ϕ, (3i^−5j^−2k^) T for B→ and 100 cm2 k^ for A→ in equation (I).

ϕ5=(3i^−5j^−2k^) T⋅(100 cm2k^)=(3i^−5j^−2k^) T⋅(100 cm2(1 m2104 cm2))k^=−0.02 Wb

Substitute ϕ6 for ϕ, (3i^−5j^−2k^) T for B→ and −100 cm2 k^ for A→ in equation (I).

ϕ6=(3i^−5j^−2k^) T⋅(−100 cm2k^)=(3i^−5j^−2k^) T⋅(−100 cm2(1 m2104 cm2))k^=0.02 Wb

Substitute 0.03 T⋅m for ϕ1 , −0.03 T⋅m for ϕ2 , −0.05 T⋅m for ϕ3 , 0.05 T⋅m for ϕ4 , −0.02 T⋅m for ϕ5 and 0.02 T⋅m for ϕ6 .

ϕnet=[(0.03 Wb)−(0.03 Wb)−(0.05 Wb)+(0.05 Wb)−(0.02 Wb)+(0.02 Wb)]=0 Wb

Thus, the net magnetic flux through all the six faces of the cube is 0 Wb.

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