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 30, Problem 76PQ

(a)

To determine

The magnitude and direction of the magnetic field at point A.

(a)

Expert Solution
Check Mark

Answer to Problem 76PQ

The magnitude of magnetic field at point A is 2.4×106T_ and the direction is positive X-axis.

Explanation of Solution

Write the expression for the magnetic field due to a current carrying wire as.

  B=(μ02π)Ir                                                                                                   (I)

Here, B is the magnetic field, I is the current flowing and r is the distance of the point from the wire.

The direction of magnetic field on the point is given by the Right hand rule. When thumb is kept in the direction of the current then the direction of curling of the finger gives the direction of the magnetic field at that point.

The magnetic field on the point A due to the wire having current I1 is in negative X-axis direction and the wire having current I2 is in positive X-axis direction.

Write the expression for the net field at the point A as.

  Bnet=B2B1                                                                                                (II)

Here, Bnet is the net magnetic field, B2  is the field due to wire 2 , and B1 is the field due to wire 1.

Conclusion:

Substitute 4π×107H/m for μ0, B1 for B,2.00A for I and 0.5m for r in equation (I).

  B1=(μ02π)2.00A0.5m(i^)=(4π×107H/m2π)2.00A0.5m(i^)=8×107T(i^)

Substitute 4π×107H/m for μ0, B2 for B,8.00A for I and 0.5m for r in equation (I).

  B2=(4π×107H/m2π)8.00A0.5m(i^)=3.2×106T(i^)

Substitute 3.2×106T(i^) for B2 and 8×107T(i^) for B1 in equation (II).

  Bnet=B2+B1=(3.2×106T)(i^)+(8×107T)(i^)=2.4×106(i^)

Thus, the magnitude of magnetic field at point A is 2.4×106T_ and the direction is positive X-axis.

(b)

To determine

The magnitude and direction of the magnetic field at point B.

(b)

Expert Solution
Check Mark

Answer to Problem 76PQ

The magnitude of magnetic field at point B is 1.44μT and direction is 56.3° below the horizontal.

Explanation of Solution

The magnetic field due to a current carrying wire is given by.

  B=(μ02π)Ir                                                                                                 (III)

Here, B is the magnetic field, I is the current flowing and r is the distance of the point from the wire.

The direction of magnetic field on the point is given by the simple Right hand rule. When the thumb is kept in the direction of the current and the palm faces the point then the direction of curling of the finger gives the direction of the magnetic field at that point.

The magnetic field on the point B due to the wire having current I1 is in negative Y-axis direction and the wire having current I2 is at 45 from the positive X-axis direction.

The field B2 is resolved as.

Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics, Chapter 30, Problem 76PQ

The net field at the point B is given as.

  Bnet=Bx2+By2                                                                                         (IV)

Here, Bx is the field at the point B due to x axis and By is the field at the point B due to y axis.

The direction of the magnetic field at point B is,

  θ=tan1(|By|Bx)                                                                                          (V)

Conclusion:

Substitute 4π×107Tm/A for μ0, B1 for B, 2.00A for I and 1.00m for r in equation (III).

  B1=(4π×107Tm/A2π)2.00A1m=0.400×106T(106μT1T)=0.400μT

Substitute 4π×107Tm/A for μ0, B2 for B, 8.00A for I and (1.00m)2 for r in equation (III).

  B2=(4π×107Tm/A2π)8.00A(1.00m)2=1.13×106T(106μT1T)=1.13μT

The net field at point B is resolved into following components.

  Bx=B2cos45By=B1B2cos45                                                                          (VI)

Substitute 1.13μT for B2 and 0.400μT for B1 in equation (VI).

  Bx=(1.13μT)cos45=0.800μTBy=(0.400μT)(1.13μT)cos45=1.20μT

Substitute 0.800μT for Bx and 1.20μT for By in equation (IV) to find Bnet.

  Bnet=(0.800μT)2+(1.20μT)2=1.44μT

Substitute 0.800μT for Bx and 1.20μT for |By| in equation (V) to find θ.

  θ=tan1(1.20μT0.800μT)=56.3° below the horizontal

Thus, the magnitude of magnetic field at point B is 1.44μT and direction is 56.3° below the horizontal.

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Chapter 30 Solutions

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

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