Physics for Scientists and Engineers with Modern Physics, Technology Update
Physics for Scientists and Engineers with Modern Physics, Technology Update
9th Edition
ISBN: 9781305401969
Author: SERWAY, Raymond A.; Jewett, John W.
Publisher: Cengage Learning
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Chapter 30, Problem 75CP

(a)

To determine

The magnitude and direction of magnetic field at point P1.

(a)

Expert Solution
Check Mark

Answer to Problem 75CP

The magnetic field at point P1 is μ0Iπr(2r2a24r2a2) directed to the left.

Explanation of Solution

Write the expression for the current density.

  J=IA                                                                                                                  (I)

Here, J is the current density, I is the current in the conductor and A is the area of conductor.

Write the expression for the area of the conductor.

  A=π(a2a24a24)=πa22                                                                                         (II)

Substitute πa22 for A in equation (I) to find J.

  J=2Iπa2                                                                                                            (III)

Write the expression for the magnetic field in the solid portion.

  Bsolid=μ0J(πa2)2πr                                                                                           (IV)

Here, Bsolid is the magnetic field in the solid portion and μ0 is the permeability of the free space.

Write the expression for the magnetic field at point 1.

  B1=μ0Jπ(a/2)22π(r(a/2))                                                                                          (V)

Write the expression for the magnetic field at point 2.

  B2=μ0Jπ(a/2)22π(r+(a/2))                                                                                         (VI)

Write the expression of net magnetic field at point P1.

  B=BsolidB1B2                                                                                            (VII)

Here, B is the net magnetic field at point P1.

Substitute μ0J(πa2)2πr for Bsolid, μ0Jπ(a/2)22π(r(a/2)) for B1 and μ0Jπ(a/2)22π(r+(a/2)) for B2 in equation (VII) to find B.

    B=μ0J(πa2)2πrμ0Jπ(a/2)22π(r(a/2))μ0Jπ(a/2)22π(r+(a/2))                                       (VIII)

Substitute 2Iπa2 for J in equation (VIII) to find B.

  B=μ0(2I)2π(4r2a22r24r(r2(a2/4)))=μ0Iπr(2r2a24r2a2)

Conclusion:

Therefore, the magnetic field at point P1 is μ0Iπr(2r2a24r2a2) directed to the left.

(b)

To determine

The magnitude and direction of magnetic field at point P2.

(b)

Expert Solution
Check Mark

Answer to Problem 75CP

The magnetic field at point P2 is μ0Iπr(2r2+a24r2+a2) towards the top of the page.

Explanation of Solution

Write the expression for the magnetic field at point 1 and point 2.

  B1=μ0Jπ(a/2)22πr2+(a/2)2=B2                                                                                      (IX)

Write the expression for the net magnetic field at point P2.

  B=BsolidB1cosθB2cosθ                                                                              (X)

Substitute μ0J(πa2)2πr for Bsolid, μ0Jπ(a/2)22πr2+(a/2)2 for B1 and B2, and (rr2+(a2/4)) for cosθ in equation (X) to find B.

  B=μ0J(πa2)2πr2[μ0Jπ(a/2)22πr2+(a/2)2](rr2+(a2/4))=μ0J(πa2)2πr[1r22(r2+(a2/4))]=2μ0I2πr[12r24r2+a2]=μ0Iπr(2r2+a24r2+a2)

Conclusion:

Therefore, the magnetic field at point P2 is μ0Iπr(2r2+a24r2+a2) towards the top of the page.

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Chapter 30 Solutions

Physics for Scientists and Engineers with Modern Physics, Technology Update

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