Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
9th Edition
ISBN: 9781305372337
Author: Raymond A. Serway | John W. Jewett
Publisher: Cengage Learning
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Chapter 30, Problem 15P

(a)

To determine

The magnitude and direction of magnetic field at point A.

(a)

Expert Solution
Check Mark

Answer to Problem 15P

The magnitude is 53.3μT and direction of magnetic field is towards the bottom of the page at point A.

Explanation of Solution

Write the expression to obtain the magnetic field along the conductor.

    B=μ0i2πr                                                                                                               (I)

Here, B is the magnetic field, μ0 is the magnetic permittivity, i is the current and r is the distance between the conductor and the point where the magnetic field is to be determined.

The arrangement of the conductors is as shown in the figure below.

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University, Chapter 30, Problem 15P

Figure-(1)

Write the expression to obtain the magnetic field at point A.

    BA=(B1+B2)cosθ+B3                                                                                      (II)

Here, BA is the magnetic field at point A, B1 is the magnetic field at point A due to conductor 1, B2 is the magnetic field at point A due to conductor 2 and B3 is the magnetic field at point A due to conductor 3.

Conclusion:

Substitute 2.00A for i, 4π×107T.m/A for μ0 and a2 for r in equation (I) to calculate B1.

    B1=(4π×107T.m/A)(2.00A)2π(a2)

Further substitute 1.00cm for a in the above equation.

    B1=(4π×107T.m/A)(2.00A)2π((1.00cm)2)=(4π×107T.m/A)(2.00A)2π((1.00cm×1m100cm)2)=282.88×107T×106μT1T=28μT

Substitute 2.00A for i, 4π×107T.m/A for μ0 and a2 for r in equation (I) to calculate B2.

    B2=(4π×107T.m/A)(2.00A)2π(a2)

Further substitute 1.00cm for a in the above equation.

    B2=(4π×107T.m/A)(2.00A)2π((1.00cm)2)=(4π×107T.m/A)(2.00A)2π((1.00cm×1m100cm)2)=282.88×107T×106μT1T=28μT

Substitute 2.00A for i, 4π×107T.m/A for μ0 and 3a for r in equation (I) to calculate B3.

    B3=(4π×107T.m/A)(2.00A)2π(3a)

Further substitute 1.00cm for a in the above equation.

    B3=(4π×107T.m/A)(2.00A)2π(3(1.00cm))=(4π×107T.m/A)(2.00A)2π(3(1.00cm×1m100cm))=133.33×107T×106μT1T=13μT

Substitute 28μT for B1 and B2, 13μT for B3 and 45° for θ in equation (II) to calculate BA.

    BA=(28μT+28μT)cos45°+13μT=(54μT)(12)+13μT=53.3μT

Based on right hand thumb rule the direction of magnetic field at point A is toward the bottom of the page.

Therefore, the magnitude is 53.3μT and direction of magnetic field is towards the bottom of the page at point A.

(b)

To determine

The magnitude and direction of magnetic field at point B.

(b)

Expert Solution
Check Mark

Answer to Problem 15P

The magnitude is 20.0μT and direction of magnetic field is towards the bottom of the page at point B.

Explanation of Solution

Write the expression to obtain the magnetic field at point A.

    BB=B3                                                                                       (III)

Here, BB is the magnetic field at point B, and B3 is the magnetic field at point B due to conductor 3.

The magnetic field due to conductor 1 cancel out the magnetic field due to conductor 2 at point B.

Conclusion:

Substitute 2.00A for i, 4π×107T.m/A for μ0 and 2a for r in equation (I) to calculate B3.

    B3=(4π×107T.m/A)(2.00A)2π(2a)

Further substitute 1.00cm for a in the above equation.

    B3=(4π×107T.m/A)(2.00A)2π(2(1.00cm))=(4π×107T.m/A)(2.00A)2π(2(1.00cm×1m100cm))=200×107T×106μT1T=20.0μT

Substitute 20.0μT for B3 and in equation (III) to calculate BB.

    BB=20.0μT

Based on right hand thumb rule the direction of magnetic field at point B is toward the bottom of the page.

Therefore, the magnitude is 20.0μT and direction of magnetic field is towards the bottom of the page at point B.

(c)

To determine

The magnitude and direction of magnetic field at point C.

(c)

Expert Solution
Check Mark

Answer to Problem 15P

The magnitude is 0 and there is no direction at point C.

Explanation of Solution

Write the expression to obtain the magnetic field at point C.

    BC=(B1+B2)sinθB3                                                                                       (IV)

Here, BC is the magnetic field at point C, B1 is the magnetic field at point C due to conductor 1, B2 is the magnetic field at point C due to conductor 2 and B3 is the magnetic field at point C due to conductor 3.

Conclusion:

Substitute 2.00A for i, 4π×107T.m/A for μ0 and a2 for r in equation (I) to calculate B1.

    B1=(4π×107T.m/A)(2.00A)2π(a2)

Further substitute 1.00cm for a in the above equation.

    B1=(4π×107T.m/A)(2.00A)2π((1.00cm)2)=(4π×107T.m/A)(2.00A)2π((1.00cm×1m100cm)2)=282.88×107T×106μT1T=28μT

Substitute 2.00A for i, 4π×107T.m/A for μ0 and a2 for r in equation (I) to calculate B2.

    B2=(4π×107T.m/A)(2.00A)2π(a2)

Further substitute 1.00cm for a in the above equation.

    B2=(4π×107T.m/A)(2.00A)2π((1.00cm)2)=(4π×107T.m/A)(2.00A)2π((1.00cm×1m100cm)2)=282.88×107T×106μT1T=28μT

Substitute 2.00A for i, 4π×107T.m/A for μ0 and a for r in equation (I) to calculate B3.

    B3=(4π×107T.m/A)(2.00A)2π(a)

Further substitute 1.00cm for a in the above equation.

    B3=(4π×107T.m/A)(2.00A)2π((1.00cm))=(4π×107T.m/A)(2.00A)2π(1.00cm×1m100cm)=400.33×107T×106μT1T=40μT

Substitute 28μT for B1 and B2, 40μT for B3 and 45° for θ in equation (II) to calculate BA.

    BC=(28μT+28μT)sin45°+40μT=(54μT)(12)40μT=39.60μT40μT40μT40μT

Further solve the above equation.

    BC=0

Therefore, the magnitude is 0 and there is no direction at point C.

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Chapter 30 Solutions

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University

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