EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
16th Edition
ISBN: 8220100546716
Author: Katz
Publisher: CENGAGE L
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Chapter 31, Problem 58PQ
To determine

The net current through the loop that must be present.

Expert Solution & Answer
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Answer to Problem 58PQ

The net current through the loop that must be present is 124.34A_.

Explanation of Solution

Write the expression for the Ampere circuital law as.

    Bdl=μ0I

Rearrange the above equation for I.

  I=B.dlμ0                                                                                                     (I)

Here, B is the magnetic field , l is length of the wire, μ0 is the permeability of free space and I is the current flowing through the wire.

The contribution to the circulation integral due to each segment of loop is.

  Circulation integral=Bdl                                                                        (II)

The two segments of the square loop are parallel to the magnetic field. Therefore, the contribution to circulation integral due to these two segments is zero.

Write the expression for the total contribution to the circulation integral as.

  Bdl=(Bdl)x+(Bdl)y                                                              (III)

Here, (Bdl)x is the circulation integral in the x-direction, (Bdl)y is the circulation integral in the y-direction and Bdl is the total circulation integral.

Conclusion:

The length of the segment in the X-direction is 2.50cm.

Substitute (Bdl)x for Bdl , (0.125yi^0.125xj^)T for B and dxi^ for dl in equation (II).

  (Bdl)x={(0.125yi^0.125xj^)Tdxi^}=0.125y02.50dx=0.125y[x]02.50=0.125y(2.50cm)

Substitute 2.50cm for y in above equation.

  (Bdl)x=0.125(2.50cm(1m100cm))(2.50cm(1m100cm))=7.8125×105Tm

The length of the segment in the Y-direction is 2.50cm.

Substitute (Bdl)y for Bdl , (0.125yi^0.125xj^)T for B and dyj^ for dl in equation (II).

  (Bdl)y={(0.125yi^0.125xj^)Tdyj^}=0.125x02.50dy=0.125x[y]02.50=0.125x(2.50 cm)

Substitute 2.50cm for x in above equation.

    (Bdl)y=0.125(2.50cm(1m100cm))(2.50cm(1m100cm))=7.8125×105Tm

Substitute 7.8125×105Tm for (Bdl)x and 7.8125×105Tm(Bdl)y in equation (III).

    Bdl=(7.8125×105Tm)+(7.8125×105Tm)=1.5625×104Tm

Substitute 1.5625×104Tm for Bdl and 4π×107TmA for μ0 in equation (I).

  I=(1.5625×104Tm)(4π×107TmA)=124.34A

Thus, the net current through the loop that must be present is 124.34A_.

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Chapter 31 Solutions

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