EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
1st Edition
ISBN: 9780100546714
Author: Katz
Publisher: YUZU
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Question
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Chapter 30, Problem 28PQ

(a)

To determine

The magnitude and direction of magnetic field at point A.

(a)

Expert Solution
Check Mark

Answer to Problem 28PQ

The magnetic field at the point A is the magnetic field at the point A is 1.00×105i^T .

Explanation of Solution

Refer to figure P30.28; the direction of the magnetic field can be described by the Right hand rule.

The Right Hand Thumb rule states that when a conductor is held such that the thumb points in the direction of the current flowing through the conductor, then the direction of curling of the fingers around the conductor shows the direction of magnetic field.

Write the expression for magnetic field at point A due to the wire carrying current I1  as.

  B1=(μ0I12πd)(i^)                                                                                           (I)

Here, d is the distance of the point A from the wire carrying current I1 , I1 is the current flowing through the conductor and the field at point A due to the wire carrying current I1  is directed towards negative X-axis.

Write the expression for magnetic field at point A due to the wire carrying current I2  as.

  B2=(μ0I22πd)(j^)                                                                                         (II)

Here, d is the distance of the point A from the wire carrying current I2 , I2 is the current flowing through the conductor and the field at point A due to the wire carrying current I2 is directed towards negative Y-axis.

Write the expression for magnetic field at point A due to the wire carrying current I3  as.

  B3=(μ0I32πd)(j^)                                                                                         (III)

Here, d is the distance of the point A from the wire carrying current I3 , I3 is the current flowing through the conductor and the field at point A due to the wire carrying current I3  is directed towards positive Y-axis.

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

  Bnet=B1+B2+B3                                                                                      (IV)

Here, Bnet is the net magnetic field.

Conclusion:

Substitute (μ0I12πd)(i^) for B1, (μ0I22πd)(j^) for B2, (μ0I32πd)(j^) for B3 in equation (IV).

    Bnet=(μ0I12πd)(i^)+(μ0I22πd)(j^)+(μ0I32πd)(j^)

Substitute 4π×107Tm/A for μ0, 0.1m for d and 5.00A for I1 , 5A for I2 and 5A for I3 in above equation.

  Bnet=[((4π×107Tm/A)(5.00A)2π(0.1m))(i^)+((4π×107Tm/A)(5.00A)2π(0.1m))(j^)+((4π×107Tm/A)(5.00A)2π(0.1m))(j^)]=(100×107T)(i^)=1.00×105i^T

Thus, the magnetic field at the point A is 1.00×105i^T .

(b)

To determine

The magnitude and direction of magnetic field at point B.

(b)

Expert Solution
Check Mark

Answer to Problem 28PQ

The magnetic field at the point B is 1.50×105i^T_ .

Explanation of Solution

Refer to figure P30.28;

Write the expression for magnetic field at point B due to the wire carrying current I1  as.

  B1=(μ0I12π(2d))(i^)                                                                                     (V)

Here, 2d is the distance of the point B from the wire carrying current I1 and the field at point B due to the wire carrying current I1  is directed towards negative X-axis.

Consider the distance of point B from the conductor carrying the current I2 is a.

  a=d2+d2=2d

Write the expression for magnetic field at point B due to the wire carrying current I2  as.

  B2=((μ0I22π(2d))cos45°)(i^)+((μ0I22π(2d))sin45°)(j^)                (VI)

Here, 2d is the distance of the point B from the wire carrying current I2 and the field at point B due to the wire carrying current I2  is directed towards 45 from negative X-axis to negative Y-axis.

Consider the distance of point B from the conductor carrying the current I3 is b.

  b=d2+d2=2d

Write the expression for magnetic field at point B due to the wire carrying current I3  as.

  B3=((μ0I32π(2d))sin45°)(i^)+((μ0I32π(2d))cos45°)(j^)                  (VII)

Here, 2d is the distance of the point B from the wire carrying current I3 and the field at point B due to the wire carrying current I3  is directed towards 45 from negative X-axis to positive Y-axis.

Write the expression for the net magnetic field at point B as.

  Bnet=B1+B2+B3                                                                                    (VIII)

Here, Bnet is the net magnetic field.

Conclusion:

Substitute (μ0I12π(2d))(i^) for B1, (μ0I22π(2d)cos45°)(i^)+(μ0I22π(2d)sin45°)(j^) for B2, (μ0I32π(2d)sin45°)(i^)+(μ0I32π(2d)cos45°)(j^) for B3 in equation (VIII).

  Bnet=[(μ0I12π(2d))(i^)+(μ0I22π(2d)cos45°)(i^)+(μ0I22π(2d)sin45°)(j^)+(μ0I32π(2d)sin45°)(i^)+(μ0I32π(2d)cos45°)(j^)]

Substitute 4π×107Tm/A for μ0, 0.1m for d  and 5.00A for I1 , 5.00A for I2 and 5.00A for I3 in above equation

  Bnet=[((4π×107Tm/A)5.00A2π(0.2 m))(i^)+((4π×107Tm/A)(5.00A)2π(2×0.1)mcos45°)(i^)((4π×107Tm/A)(5.00A)2π(2(0.1))msin45°)(j^)+((4π×107Tm/A)(5.00A)2π(2(0.1))msin45°)(i^)+((4π×107Tm/A)(5.00A)2π(2(0.1))msin45°)(j^)]=(3(4π×107Tm/A)5.00A2π(0.2 m))(i^)=(150×107T)(i^)=1.50×105i^T

Thus, the magnetic field at the point B is 1.50×105i^T_ .

(c)

To determine

The magnitude and direction of magnetic field at point C.

(c)

Expert Solution
Check Mark

Answer to Problem 28PQ

The magnetic field at the point C is [(1.2×105)(i)+(1.37×105)(j^)]T_.

Explanation of Solution

Refer to figure P30.28; write the expression for magnetic field at point C due to the wire carrying current I1  as.

  B1=(μ0I12πd)(j^)                                                                                          (IX)

Here, d is the distance of the point C from the wire carrying current I1 and the field at point C due to the wire carrying current I1  is directed towards positive Y-axis.

Write the expression for magnetic field at point A due to the wire carrying current I2  as.

  B2=(μ0I22πd)(i^)                                                                                            (X)

Here, d is the distance of the point C from the wire carrying current I2 , I2 is the current flowing through the conductor and the field at point C due to the wire carrying current I2  is directed towards positive X-axis.

Consider the distance of point B from the conductor carrying the current I3 is a.

  a=d2+(2d)2=5d

The angle made by the magnetic field vector from the X-axis is.

  tanθ=2ddθ=tan12=63.43°

Write the expression for magnetic field at point B due to the wire carrying current I2  as.

  B3=(μ0I32π(5d)cos63.43°)(i^)+(μ0I32π(5d)sin63.43°)(j^)                   (XI)

Here, 5d is the distance of the point C from the wire carrying current I3 and the field at point C due to the wire carrying current I3  is directed towards 63.43° from positive X-axis to positive Y-axis.

The net magnetic field at point C is given as.

  Bnet=B1+B2+B3                                                                                     (XII)

Here, Bnet is the net magnetic field.

Conclusion:

Substitute (μ0I12πd)(j^) for B1, (μ0I22πd)(i^) for B2 and (μ0I32π(5d)cos63.43°)(i^)+(μ0I32π(5d)sin63.43°)(j^) for B3 in equation (XII).

  Bnet=[(μ0I12πd)(j^)+(μ0I22πd)(i^)+(μ0I32π(5d)cos63.43°)(i^)+(μ0I32π(5d)sin63.43°)(j^)]

Substitute 4π×107Tm/A for μ0, 0.1 m for d, 5.00 A for I1, 5.00 A for I2 and 5.00 A for I3 in above equation.

  Bnet=[((4π×107Tm/A)(5.00A)2π(0.1 m))(j^)+((4π×107Tm/A)(5.00A)2π(0.1 m))(i^)+((4π×107Tm/A)(5.00A)(0.45)2π(5(0.1)) m)(i^)+((4π×107Tm/A)(5.00A)(0.84)2π(5(0.1)) m)(j^)]=[(105)(j^)+(105)(i^)+(0.2×105)(i^)+(0.37×105)(j^)]T=[(1.2×105)(i^)+(1.37×105)(j^)]T

Thus, the magnetic field at the point C is [(1.2×105)(i)+(1.37×105)(j^)]T_ .

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Chapter 30 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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