Physics for Scientists and Engineers: Foundations and Connections
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781133939146
Author: Katz, Debora M.
Publisher: Cengage Learning
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Chapter 30, Problem 77PQ

(a)

To determine

The magnitude and direction of magnetic field at point A.

(a)

Expert Solution
Check Mark

Answer to Problem 77PQ

The magnetic field at point A is 3.13×106k^T.

Explanation of Solution

The direction of the magnetic field at the point A is given by the Right Hand Rule. The thumb of the right hand is kept in the direction of the current and the palm faces the point, then the curl of the fingers give the direction of the magnetic field.

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

B1A=μ0I12πbk^ (I)

Here, B1A is the magnetic field due to current I1, μ0 is the permeability of free space, I1 is the current flowing in the wire and b is the distance of point A from the wire.

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

B2A=μ0I22πak^ (II)

Here, B2A is the magnetic field due to current I2 , μ0 is the permeability of free space, I2 is the current flowing in the wire and a is the distance of point A from the wire.

Write the expression for the net magnetic field as.

  BA=B1A+B2A                                                                                             (III)

Here, BA is the net magnetic field at point A.

Conclusion:

Substitute 4π×107TmA for μ02.00A for I1 and 50.0cm for b in equation (I).

  B1A=(4π×107TmA)(2.00A)2π(50.0cm)k^=4×107Tm(50.0cm(1m100cm))k^=(8×107T)k^

Substitute 4π×107TmA for μ07.00A for I2 and 60.0cm for a in equation (II).

  B2A=(4π×107TmA)(7.00A)2π(60.0cm)k^=14×107Tm(60.0cm(1m100cm))k^=(2.33×106T)k^

Substitute 8×107k^T for B1A and 2.33×106k^T for B2A in equation (III).

  BA=(8×107T)k^+(2.33×106T)k^=(3.13×106T)k^

Thus, the magnetic field at point A is 3.13×106k^T.

(b)

To determine

The magnitude and direction of magnetic field at point z=50cm.

(b)

Expert Solution
Check Mark

Answer to Problem 77PQ

The magnetic field at point z=50cm is (2.80×106T)i^(0.80×106T)(j^).

Explanation of Solution

Consider the point at z=50cm as point B.

The direction of the magnetic field at the point B is given by the Right Hand Rule. The thumb of the right hand is kept in the direction of the current and the palm faces the point, then the curl of the fingers give the direction of the magnetic field.

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

B1B=μ0I12πr(j^) (IV)

Here, B1B is the magnetic field due to current I1 , μ0 is the permeability of free space, I1 is the current flowing in the wire and r is the distance of point B from the wire.

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

B2B=μ0I22πri^ (V)

Here, B2B is the magnetic field due to current I2 , μ0 is the permeability of free space and I2 is the current flowing in the wire.

Write the expression for the net magnetic field as.

  BB=B1B+B2B                                                                                             (VI)

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

Conclusion:

Substitute 4π×107TmA for μ02.00A for I1 and 50.0cm for r in equation (IV).

  B1B=(4π×107TmA)(2.00A)2π(50.0cm)(j^)=4×107Tm(50.0cm(1m100cm))(j^)=(8×107T)(j^)

Substitute 4π×107TmA for μ07.00A for I2 and 50.0cm for r in equation (V).

  B2B=(4π×107TmA)(7.00A)2π(50.0cm)i^=14×107Tm(50.0cm(1m100cm))i^=(2.80×106T)i^

Substitute 8×107j^T for B1B and 2.80×106i^T for B2B in equation (III).

  BB=(8×107T)(j^)+(2.80×106T)i^=(2.80×106T)i^(0.80×106T)(j^)

Thus, the magnetic field at point z=50cm is (2.80×106T)i^(0.80×106T)(j^).

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Chapter 30 Solutions

Physics for Scientists and Engineers: Foundations and Connections

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