Physics for Scientists and Engineers, Vol. 1
Physics for Scientists and Engineers, Vol. 1
6th Edition
ISBN: 9781429201322
Author: Paul A. Tipler, Gene Mosca
Publisher: Macmillan Higher Education
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Chapter 28, Problem 26P

(a)

To determine

The magnetic flux through the coil.

(a)

Expert Solution
Check Mark

Answer to Problem 26P

The magnetic flux through the coilis 30.2mWb .

Explanation of Solution

Given:

The number of turn of coil is N=15.0 .

The uniform magnetic fieldis B=(4.00kG)i^ .

The radius of coilis r=4.00cm .

The unit normal to the plane of the coilis n^=i^ .

Formula used:

The expression for magnetic flux through coil is given by,

  ϕm=N(Bn^)πr2

Calculation:

The magnetic flux through coil is calculated as,

  ϕm=N(Bn^)πr2=(15.0)[(4.00kG i ^)i^]π(4.00cm)2=(15.0)[(( ( 4.00kG )( 10 1 T 1kG )) i ^)i^]π(( 4.00cm)( 10 2 m 1cm ))2=30.2×103Wb

Solve further as,

  ϕm=30.2×103Wb( 10 3 mWb 1Wb)=30.2mWb

Conclusion:

Therefore, the magnetic flux through the coilis 30.2mWb .

(b)

To determine

The magnetic flux through the coil.

(b)

Expert Solution
Check Mark

Answer to Problem 26P

The magnetic flux through the coilis 0 .

Explanation of Solution

Given:

The unit normal to the plane of the coilis n^=j^ .

Formula used:

The expression for magnetic flux through coil is given by,

  ϕm=N(Bn^)πr2

Calculation:

The magnetic flux through coil is calculated as,

  ϕm=N(Bn^)πr2=(15.0)[(4.00kG i ^)j^]π(4.00cm)2=(15.0)[(( ( 4.00kG )( 10 1 T 1kG )) i ^)j^]π(( 4.00cm)( 10 2 m 1cm ))2=0

Conclusion:

Therefore, the magnetic flux through the coilis 0 .

(c)

To determine

The magnetic flux through the coil.

(c)

Expert Solution
Check Mark

Answer to Problem 26P

The magnetic flux through the coilis 21.35mWb .

Explanation of Solution

Given:

The unit normal to the plane of the coilis n^=(i^+j^)/2 .

Formula used:

The expression for magnetic flux through coil is given by,

  ϕm=N(Bn^)πr2

Calculation:

The magnetic flux through coil is calculated as,

  ϕm=N(Bn^)πr2=(15.0)[(4.00kG i ^)( ( i ^ + j ^ )/ 2 )]π(4.00cm)2=(15.0)[(( ( 4.00kG )( 10 1 T 1kG )) i ^)( ( i ^ + j ^ )/ 2 )]π(( 4.00cm)( 10 2 m 1cm ))2=(15.0)[( ( ( 4.00kG )( 10 1 T 1kG ) ) i ^ ) i ^ 2+0]π(( 4.00cm)( 10 2 m 1cm ))2

Solve further as,

  ϕm=21.35×103Wb=21.35×103Wb( 10 3 mWb 1Wb)=21.35mWb

Conclusion:

Therefore, the magnetic flux through the coilis 21.35mWb .

(d)

To determine

The magnetic flux through the coil.

(d)

Expert Solution
Check Mark

Answer to Problem 26P

The magnetic flux through the coilis 0 .

Explanation of Solution

Given:

The unit normal to the plane of the coilis n^=k^ .

Formula used:

The expression for magnetic flux through coil is given by,

  ϕm=N(Bn^)πr2

Calculation:

The magnetic flux through coil is calculated as,

  ϕm=N(Bn^)πr2=(15.0)[(4.00kG i ^)k^]π(4.00cm)2=(15.0)[(( ( 4.00kG )( 10 1 T 1kG )) i ^)k^]π(( 4.00cm)( 10 2 m 1cm ))2=0

Conclusion:

Therefore, the magnetic flux through the coilis 0 .

(e)

To determine

The magnetic flux through the coil.

(e)

Expert Solution
Check Mark

Answer to Problem 26P

The magnetic flux through the coilis 18.12mWb .

Explanation of Solution

Given:

The unit normal to the plane of the coilis n^=0.60i^+0.80j^ .

Formula used:

The expression for magnetic flux through coil is given by,

  ϕm=N(Bn^)πr2

Calculation:

The magnetic flux through coil is calculated as,

  ϕm=N(Bn^)πr2=(15.0)[(4.00kG i ^)(0.60 i ^+0.80 j ^)]π(4.00cm)2=(15.0)[(( ( 4.00kG )( 10 1 T 1kG )) i ^)(0.60 i ^+0.80 j ^)]π(( 4.00cm)( 10 2 m 1cm ))2=(15.0)[(( ( 4.00kG )( 10 1 T 1kG )) i ^)(0.60 i ^)+0]π(( 4.00cm)( 10 2 m 1cm ))2

Solve further as,

  ϕm=18.12×103Wb=18.12×103Wb( 10 3 mWb 1Wb)=18.12mWb

Conclusion:

Therefore, the magnetic flux through the coilis 18.12mWb .

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Physics for Scientists and Engineers, Vol. 1

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