Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Question
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Chapter 30, Problem 30.15P

(a)

To determine

The magnitude and direction of magnetic field at point A .

(a)

Expert Solution
Check Mark

Answer to Problem 30.15P

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

Explanation of Solution

Given Info: The current flowing through the conductor is 2.00A , distance between corner of square to centre of square is 1.00cm .

Explanation:

Diagram of three parallel conductor having current of magnitude I is given below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 30, Problem 30.15P

Figure (1)

Formula to calculate side of the square is,

x=a2+a2=2a

Formula to calculate angle θ

θ=tan1(aa)=tan1(1)=45°

Formula to calculate magnetic field at point A due to conductor 1,

BA1=μoI2πr

Here

I is current flowing through conductor.

r is distance of the the point A from conductors 1.

Substitute 2a for rorx in above equation

BA1=μoI2π(2a)=μoI22πa

Formula to calculate magnetic field at point A due to conductor 2

BA2=μoI2πr

Here

I is current flowing through conductor.

r is distance of the the point A from conductors.2

Substitute 2a for rorx in above equation

BA2=μoI2π(2a)=μoI22πa

Formula to calculate magnetic field at point A due to conductor 3,

BA3=μoI2πr

Here

I is current flowing through conductor.

r is distance of the the point A from conductors 3.

Substitute 3a for rorx in above equation

BA3=μoI2π(3a)=μoI6πa

Write the expression to calculate magnetic field at point A due to conductors by summing in x-direction.

Bx=0BAxBA3+BA2cosθ+BA1cosθ=0BAx=BA3+BA2cosθ+BA1cosθ (1)

Here,

BA1 is magnetic field at A due to conductor 1.

BA2 is magnetic field at A due to conductor 2.

BA3 is magnetic field at A due to conductor 3.

Substitute μoI22πa for BA1andBA2 and μoI6πa for BA3 and 45° for θ in equation (1).

BAx=μoI6πa+μoI22πacos45°+μoI22πacos45°=μoI6πa+μoI22πa(12)+μoI22πa(12)=μoI6πa+μoI4πa+μoI4πa=23μoIπa

substitute 4π×107TmA for μo , 2.00A for I and 1.00cm for a in above equation.

BAx=23((4π×107TmA)(2.00A)π(1.00cm))=23((4π×107TmA)(2.00A)π(1.00cm102m1cm))=53.3×106T=53.3μT

Write the expression to calculate y-component of magnetic field at A by summing all in y-direction.

BY=0BAyBA1sin45°+BA2sin45°=0

Substitute μoI22πa for BA1 and μoI22πa for BA2 .

BAyμoI22πasin45+μoI22πacos45=0BAy=0

Formula to calculate net magnetic field at point A ,

BA=BAx2+BAy2 (2)

Substitute 53.3μT for BAx and 0 for BAy in above equation.

BA=(53.3μT)2+0=53.3μT

Hence, magnetic field at point A is 53.3μT and direction is toward the bottom of page from rule of right hand with grasp figure where thumb shows direction of current and grasp figure shows plane of magnetic field.

Conclusion:

Therefore magnetic field at point A is 53.3μT and direction is toward the bottom of page from rule of right hand with grasp figure where thumb shows direction of current and grasp figure shows plane of magnetic field.

(b)

To determine

magnitude and direction of magnetic field at point B .

(b)

Expert Solution
Check Mark

Answer to Problem 30.15P

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

Explanation of Solution

Formula to calculate magnetic field at point B due to conductor 1,

BB1=μoI2πr

Here

I is current flowing through conductor.

r is distance of the the point B from conductors 1.

Substitute a for rorx in above equation

BB1=μoI2π(a)=μoI2πa

Formula to calculate magnetic field at point B due to conductor 2

BB2=μoI2πr

Here

I is current flowing through conductor.

r is distance of the the point B from conductors.2

Substitute a for rorx in above equation

BB2=μoI2π(a)=μoI2πa

Formula to calculate magnetic field at point B due to conductor 3,

BB3=μoI2πr

Here

I is current flowing through conductor.

r is distance of the the point B from conductors 3.

Substitute 2a for rorx in above equation

BB3=μoI2π(2a)=μoI4πa

Write the expression to calculate magnetic field at point B due to conductors by summing in x-direction.

Bx=0BBxBB3=0BBx=BB3 (3)

Here,

BB1 is magnetic field at B due to conductor 1.

BB2 is magnetic field at B due to conductor 2.

BB3 is magnetic field at B due to conductor 3.

Substitute μoI4πa for BB3 in equation (3).

BBx=μoI4πa

substitute 4π×107TmA for μo , 2.00A for I and 1.00cm for a in above equation.

BBx=((4π×107TmA)(2.00A)4π(1.00cm))=((4π×107TmA)(2.00A)4π(1.00cm102m1cm))=20×106T=20μT

Write the expression to calculate y-component of magnetic field at B by summing all in y-direction.

BY=0BByBB1+BB2=0

Substitute μoI2πa for BB2 and μoI2πa for BB1 in above equation.

BByμoI2πa+μoI2πa=0BBy=0

Formula to calculate net magnetic field at point A ,

BB=BBx2+BBy2 (II)

Substitute 53.3μT for BAx and 0 for BAy in above equation.

BB=(20.0μT)2+0=20.0μT

Hence, magnetic field at point B is 20.0μT and direction is toward the bottom of page from rule of right hand with grasp figure where thumb shows direction of current and grasp figure shows plane of magnetic field.

Conclusion:

Therefore magnetic field at point B is 20.0μT and direction is toward the bottom of page from rule of right hand with grasp figure where thumb shows direction of current and grasp figure shows plane of magnetic field.

(c)

To determine

magnitude and direction of magnetic field at point C .

(c)

Expert Solution
Check Mark

Answer to Problem 30.15P

magnitude of magnetic field at point C is 0 .

Explanation of Solution

Formula to calculate magnetic field at point C due to conductor 1,

BC1=μoI2πr

Here

I is current flowing through conductor.

r is distance of the the point C from conductors 1.

Substitute 2a for rorx in above equation

BC1=μoI2π(2a)=μoI22πa

Formula to calculate magnetic field at point C due to conductor 2

BC2=μoI2πr

Here

I is current flowing through conductor.

r is distance of the the point C from conductors.2

Substitute 2a for rorx in above equation

BC2=μoI2π(2a)=μoI22πa

Formula to calculate magnetic field at point C due to conductor 3,

BC3=μoI2πr

Here

I is current flowing through conductor.

r is distance of the the point C from conductors 3.

Substitute a for rorx in above equation

BC3=μoI2π(a)=μoI2πa

Write the expression to calculate magnetic field at point A due to conductors by summing in x-direction.

Bx=0BCxBC3+BC2cosθ+BC1cosθ=0BCx=BC3+BC2cosθ+BC1cosθ (I)

Here,

BC1 is magnetic field at C due to conductor 1.

BC2 is magnetic field at C due to conductor 2.

BC3 is magnetic field at C due to conductor 3.

Substitute μoI22πa for BC1andBC2 and μoI2πa for BC3 in equation (I) and 45° for θ in equation (I).

BCx=μoI2πa+μoI22πacos45°+μoI22πacos45°=μoI2πa+μoI22πa(12)+μoI22πa(12)=μoI2πa+μoI4πa+μoI4πa=0

Write the expression to calculate y-component of magnetic field at C by summing all in y-direction.

BY=0BCyBC1sin45°+BC2sin45°=0

Substitute μoI22πa for BC1 and μoI22πa for BC2 .

BCyμoI22πasin45+μoI22πacos45=0BCy=0

Formula to calculate net magnetic field at point C ,

BC=BCx2+BCy2 (II)

Substitute 0μT for Bcx and 0 for BCy in above equation.

BC=0+0=0

Hence, magnetic field at point A is 0 .

Conclusion:

Therefore magnetic field at point A is 0 .

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Chapter 30 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

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