Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics
Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics
1st Edition
ISBN: 9781305259836
Author: Debora M. Katz
Publisher: Cengage Learning
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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.

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Answer to Problem 17PQ

The contribution of side 1 is 0.0313 Tm , the contribution of side 2 is 0.0438 Tm , the contribution of side 3 is 0.0313 Tm and the contribution of side 4 is 0.0438 Tm.

Explanation of Solution

Draw the figure for the system as shown below:

Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics, 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)(0.025m)i^=0.03125  Tm

 Further,simplify the above equation as.

     (μ0Ithru)AB=0.0313 Tm

 Thus, contribution of side AB is 0.0313 Tm_.

 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  Tm

 Further,simplify the above equation as.

     (μ0Ithru)BC=0.0438 Tm

 Thus, contribution of side BC is 0.0438 Tm_.

 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  Tm

 Further,simplify the above equation as.

     (μ0Ithru)CD=0.0313 Tm

 Thus, the contribution of side CD is 0.0313 Tm_.

 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)(0.025m)j^=0.0438  Tm

 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 Tm , the contribution of side 2 is 0.0438 Tm , the contribution of side 3 is 0.0313 Tm and the contribution of side 4 is 0.0438 Tm.

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Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics

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