The contribution to the circulation integral due to each segment of the loop, and the net current through the loop that must be present.

Question

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Answer 1

Question

Chapter 31, Problem 17PQ

To determine

The contribution to the circulation integral due to each segment of the loop, and the net current through the loop that must be present.

Expert Solution & Answer

Answer to Problem 17PQ

The contribution of side 1 is $0.0313 T⋅m$ , the contribution of side 2 is $−0.0438 T⋅m$ , the contribution of side 3 is $−0.0313 T⋅m$ and the contribution of side 4 is $0.0438 T⋅m$ .

Explanation of Solution

Draw the figure for the system as shown below:

Physics for Scientists and Engineers: Foundations and Connections, Chapter 31, Problem 17PQ

Refer to the above figure, consider a square loop $ABCD$ in $xy-plane$ . The sides $AB$ and $CD$ have equal magnitude in opposite direction whereas side $BC$ and side $DA$ has equal magnitude in opposite direction.

Write the general expression for Ampere’s Circuital Law for the given above figure as.

$μ0Ithru=∮B→.dl→$ (I)

Here, $dl→$ is length of element, $μ0$ is the permittivity of free space, $Ithru$ is the current through the conductor and $B→$ is magnetic field.

Substitute $dxi^$ for $dl→$ in equation (I) for side $AB$ as.

$(μ0Ithru)AB=∫ABB→.dxi^$ (II)

Substitute $dy(−j^)$ for $dl→$ in equation (I) for side $BC$ as.

$(μ0Ithru)BC=∫BCB→.(−dy)j^$ (III)

Substitute $dx(−i^)$ for $dl→$ in equation (I) for side $CD$ as.

$(μ0Ithru)CD=∫CDB→.(−dx)i^$ (IV)

Substitute $dy(j^)$ for $dl→$ in equation (I) for side $DA$ as.

$(μ0Ithru)DA=∫DAB→.dyj^$ (V)

Conclusion:

Substitute $(1.25i^+1.75j^)T$ for $B→$ and $2.50 cmi^$ for $∫dxi^$ in equation (II).

$(μ0Ithru)AB=((1.25i^+1.75j^)T)⋅(2.50 cm)i^=((1.25i^+1.75j^)T)⋅(2.50 cm(1 m100 cm))i^=((1.25i^+1.75j^)T)⋅(0.025 m)i^=0.03125 T⋅m$

Further,simplify the above equation as.

$(μ0Ithru)AB=0.0313 T⋅m$

Thus, contribution of side $AB$ is $0.0313 T⋅m_$ .

Substitute $(1.25i^+1.75j^)T$ for $B→$ and $2.50 cm(−j^)$ for $∫dy(−j^)$ in equation (III).

$(μ0Ithru)BC=((1.25i^+1.75j^)T)⋅(2.50 cm)(−j^)=((1.25i^+1.75j^)T)⋅(2.50 cm(1 m100 cm))(−j^)=((1.25i^+1.75j^)T)⋅(0.025 m)(−j^)=−0.0438 T⋅m$

Further,simplify the above equation as.

$(μ0Ithru)BC=−0.0438 T⋅m$

Thus, contribution of side $BC$ is $−0.0438 T⋅m_$ .

Substitute $(1.25i^+1.75j^) T$ for $B→$ and $2.50 cm(−i^)$ for $∫dx(−i^)$ in equation (IV).

$(μ0Ithru)CD=((1.25i^+1.75j^) T)⋅(2.50 cm)(−i^)=((1.25i^+1.75j^) T)⋅(2.50 cm×1 m100 cm)(−i^)=((1.25i^+1.75j^) T)⋅(0.025 m)(−i^)=−0.03125 T⋅m$

Further,simplify the above equation as.

$(μ0Ithru)CD=−0.0313 T⋅m$

Thus, the contribution of side $CD$ is $−0.0313 T⋅m_$ .

Substitute $(1.25i^+1.75j^) T$ for $B→$ and $(2.50 cm)j^$ for $∫dyj^$ in equation (V).

$(μ0Ithru)DA=((1.25i^+1.75j^) T)⋅(2.50 cm)j^=((1.25i^+1.75j^) T)⋅(2.50 cm(1 m100 cm))j^=((1.25i^+1.75j^) T)⋅(0.025 m)j^=0.0438 T⋅m$

Further,simplify the above equation as.

$(μ0Ithru)DA=0.0438 T$

Thus, the contribution of side $AB$ is $0.0438 T_$ .

Thus, the contribution of side 1 is $0.0313 T⋅m$ , the contribution of side 2 is $−0.0438 T⋅m$ , the contribution of side 3 is $−0.0313 T⋅m$ and the contribution of side 4 is $0.0438 T⋅m$ .

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Answer 2

Question

Chapter 31, Problem 17PQ

To determine

The contribution to the circulation integral due to each segment of the loop, and the net current through the loop that must be present.

Expert Solution & Answer

Answer to Problem 17PQ

The contribution of side 1 is $0.0313 T⋅m$ , the contribution of side 2 is $−0.0438 T⋅m$ , the contribution of side 3 is $−0.0313 T⋅m$ and the contribution of side 4 is $0.0438 T⋅m$ .

Explanation of Solution

Draw the figure for the system as shown below:

Physics for Scientists and Engineers: Foundations and Connections, Chapter 31, Problem 17PQ

Refer to the above figure, consider a square loop $ABCD$ in $xy-plane$ . The sides $AB$ and $CD$ have equal magnitude in opposite direction whereas side $BC$ and side $DA$ has equal magnitude in opposite direction.

Write the general expression for Ampere’s Circuital Law for the given above figure as.

$μ0Ithru=∮B→.dl→$ (I)

Here, $dl→$ is length of element, $μ0$ is the permittivity of free space, $Ithru$ is the current through the conductor and $B→$ is magnetic field.

Substitute $dxi^$ for $dl→$ in equation (I) for side $AB$ as.

$(μ0Ithru)AB=∫ABB→.dxi^$ (II)

Substitute $dy(−j^)$ for $dl→$ in equation (I) for side $BC$ as.

$(μ0Ithru)BC=∫BCB→.(−dy)j^$ (III)

Substitute $dx(−i^)$ for $dl→$ in equation (I) for side $CD$ as.

$(μ0Ithru)CD=∫CDB→.(−dx)i^$ (IV)

Substitute $dy(j^)$ for $dl→$ in equation (I) for side $DA$ as.

$(μ0Ithru)DA=∫DAB→.dyj^$ (V)

Conclusion:

Substitute $(1.25i^+1.75j^)T$ for $B→$ and $2.50 cmi^$ for $∫dxi^$ in equation (II).

$(μ0Ithru)AB=((1.25i^+1.75j^)T)⋅(2.50 cm)i^=((1.25i^+1.75j^)T)⋅(2.50 cm(1 m100 cm))i^=((1.25i^+1.75j^)T)⋅(0.025 m)i^=0.03125 T⋅m$

Further,simplify the above equation as.

$(μ0Ithru)AB=0.0313 T⋅m$

Thus, contribution of side $AB$ is $0.0313 T⋅m_$ .

Substitute $(1.25i^+1.75j^)T$ for $B→$ and $2.50 cm(−j^)$ for $∫dy(−j^)$ in equation (III).

$(μ0Ithru)BC=((1.25i^+1.75j^)T)⋅(2.50 cm)(−j^)=((1.25i^+1.75j^)T)⋅(2.50 cm(1 m100 cm))(−j^)=((1.25i^+1.75j^)T)⋅(0.025 m)(−j^)=−0.0438 T⋅m$

Further,simplify the above equation as.

$(μ0Ithru)BC=−0.0438 T⋅m$

Thus, contribution of side $BC$ is $−0.0438 T⋅m_$ .

Substitute $(1.25i^+1.75j^) T$ for $B→$ and $2.50 cm(−i^)$ for $∫dx(−i^)$ in equation (IV).

$(μ0Ithru)CD=((1.25i^+1.75j^) T)⋅(2.50 cm)(−i^)=((1.25i^+1.75j^) T)⋅(2.50 cm×1 m100 cm)(−i^)=((1.25i^+1.75j^) T)⋅(0.025 m)(−i^)=−0.03125 T⋅m$

Further,simplify the above equation as.

$(μ0Ithru)CD=−0.0313 T⋅m$

Thus, the contribution of side $CD$ is $−0.0313 T⋅m_$ .

Substitute $(1.25i^+1.75j^) T$ for $B→$ and $(2.50 cm)j^$ for $∫dyj^$ in equation (V).

$(μ0Ithru)DA=((1.25i^+1.75j^) T)⋅(2.50 cm)j^=((1.25i^+1.75j^) T)⋅(2.50 cm(1 m100 cm))j^=((1.25i^+1.75j^) T)⋅(0.025 m)j^=0.0438 T⋅m$

Further,simplify the above equation as.

$(μ0Ithru)DA=0.0438 T$

Thus, the contribution of side $AB$ is $0.0438 T_$ .

Thus, the contribution of side 1 is $0.0313 T⋅m$ , the contribution of side 2 is $−0.0438 T⋅m$ , the contribution of side 3 is $−0.0313 T⋅m$ and the contribution of side 4 is $0.0438 T⋅m$ .

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Answer 3

Question

Chapter 31, Problem 17PQ

To determine

The contribution to the circulation integral due to each segment of the loop, and the net current through the loop that must be present.

Expert Solution & Answer

Answer to Problem 17PQ

The contribution of side 1 is $0.0313 T⋅m$ , the contribution of side 2 is $−0.0438 T⋅m$ , the contribution of side 3 is $−0.0313 T⋅m$ and the contribution of side 4 is $0.0438 T⋅m$ .

Explanation of Solution

Draw the figure for the system as shown below:

Physics for Scientists and Engineers: Foundations and Connections, Chapter 31, Problem 17PQ

Refer to the above figure, consider a square loop $ABCD$ in $xy-plane$ . The sides $AB$ and $CD$ have equal magnitude in opposite direction whereas side $BC$ and side $DA$ has equal magnitude in opposite direction.

Write the general expression for Ampere’s Circuital Law for the given above figure as.

$μ0Ithru=∮B→.dl→$ (I)

Here, $dl→$ is length of element, $μ0$ is the permittivity of free space, $Ithru$ is the current through the conductor and $B→$ is magnetic field.

Substitute $dxi^$ for $dl→$ in equation (I) for side $AB$ as.

$(μ0Ithru)AB=∫ABB→.dxi^$ (II)

Substitute $dy(−j^)$ for $dl→$ in equation (I) for side $BC$ as.

$(μ0Ithru)BC=∫BCB→.(−dy)j^$ (III)

Substitute $dx(−i^)$ for $dl→$ in equation (I) for side $CD$ as.

$(μ0Ithru)CD=∫CDB→.(−dx)i^$ (IV)

Substitute $dy(j^)$ for $dl→$ in equation (I) for side $DA$ as.

$(μ0Ithru)DA=∫DAB→.dyj^$ (V)

Conclusion:

Substitute $(1.25i^+1.75j^)T$ for $B→$ and $2.50 cmi^$ for $∫dxi^$ in equation (II).

$(μ0Ithru)AB=((1.25i^+1.75j^)T)⋅(2.50 cm)i^=((1.25i^+1.75j^)T)⋅(2.50 cm(1 m100 cm))i^=((1.25i^+1.75j^)T)⋅(0.025 m)i^=0.03125 T⋅m$

Further,simplify the above equation as.

$(μ0Ithru)AB=0.0313 T⋅m$

Thus, contribution of side $AB$ is $0.0313 T⋅m_$ .

Substitute $(1.25i^+1.75j^)T$ for $B→$ and $2.50 cm(−j^)$ for $∫dy(−j^)$ in equation (III).

$(μ0Ithru)BC=((1.25i^+1.75j^)T)⋅(2.50 cm)(−j^)=((1.25i^+1.75j^)T)⋅(2.50 cm(1 m100 cm))(−j^)=((1.25i^+1.75j^)T)⋅(0.025 m)(−j^)=−0.0438 T⋅m$

Further,simplify the above equation as.

$(μ0Ithru)BC=−0.0438 T⋅m$

Thus, contribution of side $BC$ is $−0.0438 T⋅m_$ .

Substitute $(1.25i^+1.75j^) T$ for $B→$ and $2.50 cm(−i^)$ for $∫dx(−i^)$ in equation (IV).

$(μ0Ithru)CD=((1.25i^+1.75j^) T)⋅(2.50 cm)(−i^)=((1.25i^+1.75j^) T)⋅(2.50 cm×1 m100 cm)(−i^)=((1.25i^+1.75j^) T)⋅(0.025 m)(−i^)=−0.03125 T⋅m$

Further,simplify the above equation as.

$(μ0Ithru)CD=−0.0313 T⋅m$

Thus, the contribution of side $CD$ is $−0.0313 T⋅m_$ .

Substitute $(1.25i^+1.75j^) T$ for $B→$ and $(2.50 cm)j^$ for $∫dyj^$ in equation (V).

$(μ0Ithru)DA=((1.25i^+1.75j^) T)⋅(2.50 cm)j^=((1.25i^+1.75j^) T)⋅(2.50 cm(1 m100 cm))j^=((1.25i^+1.75j^) T)⋅(0.025 m)j^=0.0438 T⋅m$

Further,simplify the above equation as.

$(μ0Ithru)DA=0.0438 T$

Thus, the contribution of side $AB$ is $0.0438 T_$ .

Thus, the contribution of side 1 is $0.0313 T⋅m$ , the contribution of side 2 is $−0.0438 T⋅m$ , the contribution of side 3 is $−0.0313 T⋅m$ and the contribution of side 4 is $0.0438 T⋅m$ .

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Answer 4

Question

Chapter 31, Problem 17PQ

To determine

The contribution to the circulation integral due to each segment of the loop, and the net current through the loop that must be present.

Expert Solution & Answer

Answer to Problem 17PQ

The contribution of side 1 is $0.0313 T⋅m$ , the contribution of side 2 is $−0.0438 T⋅m$ , the contribution of side 3 is $−0.0313 T⋅m$ and the contribution of side 4 is $0.0438 T⋅m$ .

Explanation of Solution

Draw the figure for the system as shown below:

Physics for Scientists and Engineers: Foundations and Connections, Chapter 31, Problem 17PQ

Refer to the above figure, consider a square loop $ABCD$ in $xy-plane$ . The sides $AB$ and $CD$ have equal magnitude in opposite direction whereas side $BC$ and side $DA$ has equal magnitude in opposite direction.

Write the general expression for Ampere’s Circuital Law for the given above figure as.

$μ0Ithru=∮B→.dl→$ (I)

Here, $dl→$ is length of element, $μ0$ is the permittivity of free space, $Ithru$ is the current through the conductor and $B→$ is magnetic field.

Substitute $dxi^$ for $dl→$ in equation (I) for side $AB$ as.

$(μ0Ithru)AB=∫ABB→.dxi^$ (II)

Substitute $dy(−j^)$ for $dl→$ in equation (I) for side $BC$ as.

$(μ0Ithru)BC=∫BCB→.(−dy)j^$ (III)

Substitute $dx(−i^)$ for $dl→$ in equation (I) for side $CD$ as.

$(μ0Ithru)CD=∫CDB→.(−dx)i^$ (IV)

Substitute $dy(j^)$ for $dl→$ in equation (I) for side $DA$ as.

$(μ0Ithru)DA=∫DAB→.dyj^$ (V)

Conclusion:

Substitute $(1.25i^+1.75j^)T$ for $B→$ and $2.50 cmi^$ for $∫dxi^$ in equation (II).

$(μ0Ithru)AB=((1.25i^+1.75j^)T)⋅(2.50 cm)i^=((1.25i^+1.75j^)T)⋅(2.50 cm(1 m100 cm))i^=((1.25i^+1.75j^)T)⋅(0.025 m)i^=0.03125 T⋅m$

Further,simplify the above equation as.

$(μ0Ithru)AB=0.0313 T⋅m$

Thus, contribution of side $AB$ is $0.0313 T⋅m_$ .

Substitute $(1.25i^+1.75j^)T$ for $B→$ and $2.50 cm(−j^)$ for $∫dy(−j^)$ in equation (III).

$(μ0Ithru)BC=((1.25i^+1.75j^)T)⋅(2.50 cm)(−j^)=((1.25i^+1.75j^)T)⋅(2.50 cm(1 m100 cm))(−j^)=((1.25i^+1.75j^)T)⋅(0.025 m)(−j^)=−0.0438 T⋅m$

Further,simplify the above equation as.

$(μ0Ithru)BC=−0.0438 T⋅m$

Thus, contribution of side $BC$ is $−0.0438 T⋅m_$ .

Substitute $(1.25i^+1.75j^) T$ for $B→$ and $2.50 cm(−i^)$ for $∫dx(−i^)$ in equation (IV).

$(μ0Ithru)CD=((1.25i^+1.75j^) T)⋅(2.50 cm)(−i^)=((1.25i^+1.75j^) T)⋅(2.50 cm×1 m100 cm)(−i^)=((1.25i^+1.75j^) T)⋅(0.025 m)(−i^)=−0.03125 T⋅m$

Further,simplify the above equation as.

$(μ0Ithru)CD=−0.0313 T⋅m$

Thus, the contribution of side $CD$ is $−0.0313 T⋅m_$ .

Substitute $(1.25i^+1.75j^) T$ for $B→$ and $(2.50 cm)j^$ for $∫dyj^$ in equation (V).

$(μ0Ithru)DA=((1.25i^+1.75j^) T)⋅(2.50 cm)j^=((1.25i^+1.75j^) T)⋅(2.50 cm(1 m100 cm))j^=((1.25i^+1.75j^) T)⋅(0.025 m)j^=0.0438 T⋅m$

Further,simplify the above equation as.

$(μ0Ithru)DA=0.0438 T$

Thus, the contribution of side $AB$ is $0.0438 T_$ .

Thus, the contribution of side 1 is $0.0313 T⋅m$ , the contribution of side 2 is $−0.0438 T⋅m$ , the contribution of side 3 is $−0.0313 T⋅m$ and the contribution of side 4 is $0.0438 T⋅m$ .

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Answer 5

Chapter 31, Problem 17PQ

To determine

The contribution to the circulation integral due to each segment of the loop, and the net current through the loop that must be present.

Expert Solution & Answer

Answer to Problem 17PQ

The contribution of side 1 is $0.0313 T⋅m$ , the contribution of side 2 is $−0.0438 T⋅m$ , the contribution of side 3 is $−0.0313 T⋅m$ and the contribution of side 4 is $0.0438 T⋅m$ .

Explanation of Solution

Draw the figure for the system as shown below:

Physics for Scientists and Engineers: Foundations and Connections, Chapter 31, Problem 17PQ

Refer to the above figure, consider a square loop $ABCD$ in $xy-plane$ . The sides $AB$ and $CD$ have equal magnitude in opposite direction whereas side $BC$ and side $DA$ has equal magnitude in opposite direction.

Write the general expression for Ampere’s Circuital Law for the given above figure as.

$μ0Ithru=∮B→.dl→$ (I)

Here, $dl→$ is length of element, $μ0$ is the permittivity of free space, $Ithru$ is the current through the conductor and $B→$ is magnetic field.

Substitute $dxi^$ for $dl→$ in equation (I) for side $AB$ as.

$(μ0Ithru)AB=∫ABB→.dxi^$ (II)

Substitute $dy(−j^)$ for $dl→$ in equation (I) for side $BC$ as.

$(μ0Ithru)BC=∫BCB→.(−dy)j^$ (III)

Substitute $dx(−i^)$ for $dl→$ in equation (I) for side $CD$ as.

$(μ0Ithru)CD=∫CDB→.(−dx)i^$ (IV)

Substitute $dy(j^)$ for $dl→$ in equation (I) for side $DA$ as.

$(μ0Ithru)DA=∫DAB→.dyj^$ (V)

Conclusion:

Substitute $(1.25i^+1.75j^)T$ for $B→$ and $2.50 cmi^$ for $∫dxi^$ in equation (II).

$(μ0Ithru)AB=((1.25i^+1.75j^)T)⋅(2.50 cm)i^=((1.25i^+1.75j^)T)⋅(2.50 cm(1 m100 cm))i^=((1.25i^+1.75j^)T)⋅(0.025 m)i^=0.03125 T⋅m$

Further,simplify the above equation as.

$(μ0Ithru)AB=0.0313 T⋅m$

Thus, contribution of side $AB$ is $0.0313 T⋅m_$ .

Substitute $(1.25i^+1.75j^)T$ for $B→$ and $2.50 cm(−j^)$ for $∫dy(−j^)$ in equation (III).

$(μ0Ithru)BC=((1.25i^+1.75j^)T)⋅(2.50 cm)(−j^)=((1.25i^+1.75j^)T)⋅(2.50 cm(1 m100 cm))(−j^)=((1.25i^+1.75j^)T)⋅(0.025 m)(−j^)=−0.0438 T⋅m$

Further,simplify the above equation as.

$(μ0Ithru)BC=−0.0438 T⋅m$

Thus, contribution of side $BC$ is $−0.0438 T⋅m_$ .

Substitute $(1.25i^+1.75j^) T$ for $B→$ and $2.50 cm(−i^)$ for $∫dx(−i^)$ in equation (IV).

$(μ0Ithru)CD=((1.25i^+1.75j^) T)⋅(2.50 cm)(−i^)=((1.25i^+1.75j^) T)⋅(2.50 cm×1 m100 cm)(−i^)=((1.25i^+1.75j^) T)⋅(0.025 m)(−i^)=−0.03125 T⋅m$

Further,simplify the above equation as.

$(μ0Ithru)CD=−0.0313 T⋅m$

Thus, the contribution of side $CD$ is $−0.0313 T⋅m_$ .

Substitute $(1.25i^+1.75j^) T$ for $B→$ and $(2.50 cm)j^$ for $∫dyj^$ in equation (V).

$(μ0Ithru)DA=((1.25i^+1.75j^) T)⋅(2.50 cm)j^=((1.25i^+1.75j^) T)⋅(2.50 cm(1 m100 cm))j^=((1.25i^+1.75j^) T)⋅(0.025 m)j^=0.0438 T⋅m$

Further,simplify the above equation as.

$(μ0Ithru)DA=0.0438 T$

Thus, the contribution of side $AB$ is $0.0438 T_$ .

Thus, the contribution of side 1 is $0.0313 T⋅m$ , the contribution of side 2 is $−0.0438 T⋅m$ , the contribution of side 3 is $−0.0313 T⋅m$ and the contribution of side 4 is $0.0438 T⋅m$ .

Answer 6

Chapter 31, Problem 17PQ

To determine

The contribution to the circulation integral due to each segment of the loop, and the net current through the loop that must be present.

Expert Solution & Answer

Answer to Problem 17PQ

The contribution of side 1 is $0.0313 T⋅m$ , the contribution of side 2 is $−0.0438 T⋅m$ , the contribution of side 3 is $−0.0313 T⋅m$ and the contribution of side 4 is $0.0438 T⋅m$ .

Explanation of Solution

Draw the figure for the system as shown below:

Physics for Scientists and Engineers: Foundations and Connections, Chapter 31, Problem 17PQ

Refer to the above figure, consider a square loop $ABCD$ in $xy-plane$ . The sides $AB$ and $CD$ have equal magnitude in opposite direction whereas side $BC$ and side $DA$ has equal magnitude in opposite direction.

Write the general expression for Ampere’s Circuital Law for the given above figure as.

$μ0Ithru=∮B→.dl→$ (I)

Here, $dl→$ is length of element, $μ0$ is the permittivity of free space, $Ithru$ is the current through the conductor and $B→$ is magnetic field.

Substitute $dxi^$ for $dl→$ in equation (I) for side $AB$ as.

$(μ0Ithru)AB=∫ABB→.dxi^$ (II)

Substitute $dy(−j^)$ for $dl→$ in equation (I) for side $BC$ as.

$(μ0Ithru)BC=∫BCB→.(−dy)j^$ (III)

Substitute $dx(−i^)$ for $dl→$ in equation (I) for side $CD$ as.

$(μ0Ithru)CD=∫CDB→.(−dx)i^$ (IV)

Substitute $dy(j^)$ for $dl→$ in equation (I) for side $DA$ as.

$(μ0Ithru)DA=∫DAB→.dyj^$ (V)

Conclusion:

Substitute $(1.25i^+1.75j^)T$ for $B→$ and $2.50 cmi^$ for $∫dxi^$ in equation (II).

$(μ0Ithru)AB=((1.25i^+1.75j^)T)⋅(2.50 cm)i^=((1.25i^+1.75j^)T)⋅(2.50 cm(1 m100 cm))i^=((1.25i^+1.75j^)T)⋅(0.025 m)i^=0.03125 T⋅m$

Further,simplify the above equation as.

$(μ0Ithru)AB=0.0313 T⋅m$

Thus, contribution of side $AB$ is $0.0313 T⋅m_$ .

Substitute $(1.25i^+1.75j^)T$ for $B→$ and $2.50 cm(−j^)$ for $∫dy(−j^)$ in equation (III).

$(μ0Ithru)BC=((1.25i^+1.75j^)T)⋅(2.50 cm)(−j^)=((1.25i^+1.75j^)T)⋅(2.50 cm(1 m100 cm))(−j^)=((1.25i^+1.75j^)T)⋅(0.025 m)(−j^)=−0.0438 T⋅m$

Further,simplify the above equation as.

$(μ0Ithru)BC=−0.0438 T⋅m$

Thus, contribution of side $BC$ is $−0.0438 T⋅m_$ .

Substitute $(1.25i^+1.75j^) T$ for $B→$ and $2.50 cm(−i^)$ for $∫dx(−i^)$ in equation (IV).

$(μ0Ithru)CD=((1.25i^+1.75j^) T)⋅(2.50 cm)(−i^)=((1.25i^+1.75j^) T)⋅(2.50 cm×1 m100 cm)(−i^)=((1.25i^+1.75j^) T)⋅(0.025 m)(−i^)=−0.03125 T⋅m$

Further,simplify the above equation as.

$(μ0Ithru)CD=−0.0313 T⋅m$

Thus, the contribution of side $CD$ is $−0.0313 T⋅m_$ .

Substitute $(1.25i^+1.75j^) T$ for $B→$ and $(2.50 cm)j^$ for $∫dyj^$ in equation (V).

$(μ0Ithru)DA=((1.25i^+1.75j^) T)⋅(2.50 cm)j^=((1.25i^+1.75j^) T)⋅(2.50 cm(1 m100 cm))j^=((1.25i^+1.75j^) T)⋅(0.025 m)j^=0.0438 T⋅m$

Further,simplify the above equation as.

$(μ0Ithru)DA=0.0438 T$

Thus, the contribution of side $AB$ is $0.0438 T_$ .

Thus, the contribution of side 1 is $0.0313 T⋅m$ , the contribution of side 2 is $−0.0438 T⋅m$ , the contribution of side 3 is $−0.0313 T⋅m$ and the contribution of side 4 is $0.0438 T⋅m$ .

Answer 7

Chapter 31, Problem 17PQ

To determine

The contribution to the circulation integral due to each segment of the loop, and the net current through the loop that must be present.

Answer 8

Expert Solution & Answer

Answer to Problem 17PQ

The contribution of side 1 is $0.0313 T⋅m$ , the contribution of side 2 is $−0.0438 T⋅m$ , the contribution of side 3 is $−0.0313 T⋅m$ and the contribution of side 4 is $0.0438 T⋅m$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Draw the figure for the system as shown below:

Physics for Scientists and Engineers: Foundations and Connections, Chapter 31, Problem 17PQ

Refer to the above figure, consider a square loop $ABCD$ in $xy-plane$ . The sides $AB$ and $CD$ have equal magnitude in opposite direction whereas side $BC$ and side $DA$ has equal magnitude in opposite direction.

Write the general expression for Ampere’s Circuital Law for the given above figure as.

$μ0Ithru=∮B→.dl→$ (I)

Here, $dl→$ is length of element, $μ0$ is the permittivity of free space, $Ithru$ is the current through the conductor and $B→$ is magnetic field.

Substitute $dxi^$ for $dl→$ in equation (I) for side $AB$ as.

$(μ0Ithru)AB=∫ABB→.dxi^$ (II)

Substitute $dy(−j^)$ for $dl→$ in equation (I) for side $BC$ as.

$(μ0Ithru)BC=∫BCB→.(−dy)j^$ (III)

Substitute $dx(−i^)$ for $dl→$ in equation (I) for side $CD$ as.

$(μ0Ithru)CD=∫CDB→.(−dx)i^$ (IV)

Substitute $dy(j^)$ for $dl→$ in equation (I) for side $DA$ as.

$(μ0Ithru)DA=∫DAB→.dyj^$ (V)

Conclusion:

Substitute $(1.25i^+1.75j^)T$ for $B→$ and $2.50 cmi^$ for $∫dxi^$ in equation (II).

$(μ0Ithru)AB=((1.25i^+1.75j^)T)⋅(2.50 cm)i^=((1.25i^+1.75j^)T)⋅(2.50 cm(1 m100 cm))i^=((1.25i^+1.75j^)T)⋅(0.025 m)i^=0.03125 T⋅m$

Further,simplify the above equation as.

$(μ0Ithru)AB=0.0313 T⋅m$

Thus, contribution of side $AB$ is $0.0313 T⋅m_$ .

Substitute $(1.25i^+1.75j^)T$ for $B→$ and $2.50 cm(−j^)$ for $∫dy(−j^)$ in equation (III).

$(μ0Ithru)BC=((1.25i^+1.75j^)T)⋅(2.50 cm)(−j^)=((1.25i^+1.75j^)T)⋅(2.50 cm(1 m100 cm))(−j^)=((1.25i^+1.75j^)T)⋅(0.025 m)(−j^)=−0.0438 T⋅m$

Further,simplify the above equation as.

$(μ0Ithru)BC=−0.0438 T⋅m$

Thus, contribution of side $BC$ is $−0.0438 T⋅m_$ .

Substitute $(1.25i^+1.75j^) T$ for $B→$ and $2.50 cm(−i^)$ for $∫dx(−i^)$ in equation (IV).

$(μ0Ithru)CD=((1.25i^+1.75j^) T)⋅(2.50 cm)(−i^)=((1.25i^+1.75j^) T)⋅(2.50 cm×1 m100 cm)(−i^)=((1.25i^+1.75j^) T)⋅(0.025 m)(−i^)=−0.03125 T⋅m$

Further,simplify the above equation as.

$(μ0Ithru)CD=−0.0313 T⋅m$

Thus, the contribution of side $CD$ is $−0.0313 T⋅m_$ .

Substitute $(1.25i^+1.75j^) T$ for $B→$ and $(2.50 cm)j^$ for $∫dyj^$ in equation (V).

$(μ0Ithru)DA=((1.25i^+1.75j^) T)⋅(2.50 cm)j^=((1.25i^+1.75j^) T)⋅(2.50 cm(1 m100 cm))j^=((1.25i^+1.75j^) T)⋅(0.025 m)j^=0.0438 T⋅m$

Further,simplify the above equation as.

$(μ0Ithru)DA=0.0438 T$

Thus, the contribution of side $AB$ is $0.0438 T_$ .

Thus, the contribution of side 1 is $0.0313 T⋅m$ , the contribution of side 2 is $−0.0438 T⋅m$ , the contribution of side 3 is $−0.0313 T⋅m$ and the contribution of side 4 is $0.0438 T⋅m$ .

Answer 12

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The contribution to the circulation integral due to each segment of the loop, and the net current through the loop that must be present.

Concept explainers

The contribution to the circulation integral due to each segment of the loop, and the net current through the loop that must be present.

Answer to Problem 17PQ

Explanation of Solution

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