The magnetic field at y = − 3.0 cm .

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 27, Problem 25P

(a)

To determine

The magnetic field at y=−3.0 cm .

(a)

Expert Solution

Answer to Problem 25P

The magnetic field is −89 μT .

Explanation of Solution

Formula used:

The expression for magnetic field is given by,

B=−μ0I2πR1k^+μ0I2πR2k^

Calculation:

The magnetic field is calculated as,

B=−μ0I2πR1k^+μ0I2πR2k^=−( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 3.0 cm )( 10 −2 m 1 cm ))k^+( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 9.0 cm )( 10 −2 m 1 cm ))k^=−(( 89× 10 −6 T)( 10 6 μT 1 T ))k^=−89 μT

Conclusion:

Therefore, the magnetic fieldis −89 μT .

(b)

To determine

The magnetic field at y=0 .

(b)

Expert Solution

Answer to Problem 25P

The magnetic field is 0 T .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B=−μ0I2πR1k^+μ0I2πR2k^=−( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 6.0 cm )( 10 −2 m 1 cm ))k^+( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 6.0 cm )( 10 −2 m 1 cm ))k^=0 T

Conclusion:

Therefore, the magnetic field is 0 T .

(c)

To determine

The magnetic field at y=3.0 cm .

(c)

Expert Solution

Answer to Problem 25P

The magnetic field is 89 μT .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B=−μ0I2πR1k^+μ0I2πR2k^=−( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 9.0 cm )( 10 −2 m 1 cm ))k^+( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 3.0 cm )( 10 −2 m 1 cm ))k^=(( 89× 10 −6 T)( 10 6 μT 1 T ))k^=89 μT

Conclusion:

Therefore, the magnetic field is 89 μT .

(d)

To determine

The magnetic field at y=+9.0 cm .

(d)

Expert Solution

Answer to Problem 25P

The magnetic field is −160 μT .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B=−μ0I2πR1k^+μ0I2πR2k^=−( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 15.0 cm )( 10 −2 m 1 cm ))k^+( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 3.0 cm )( 10 −2 m 1 cm ))k^=−(( 160× 10 −6 T)( 10 6 μT 1 T ))k^=−160 μT

Conclusion:

Therefore, the magnetic field is −160 μT .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 2

Question

Chapter 27, Problem 25P

(a)

To determine

The magnetic field at y=−3.0 cm .

(a)

Expert Solution

Answer to Problem 25P

The magnetic field is −89 μT .

Explanation of Solution

Formula used:

The expression for magnetic field is given by,

B=−μ0I2πR1k^+μ0I2πR2k^

Calculation:

The magnetic field is calculated as,

B=−μ0I2πR1k^+μ0I2πR2k^=−( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 3.0 cm )( 10 −2 m 1 cm ))k^+( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 9.0 cm )( 10 −2 m 1 cm ))k^=−(( 89× 10 −6 T)( 10 6 μT 1 T ))k^=−89 μT

Conclusion:

Therefore, the magnetic fieldis −89 μT .

(b)

To determine

The magnetic field at y=0 .

(b)

Expert Solution

Answer to Problem 25P

The magnetic field is 0 T .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B=−μ0I2πR1k^+μ0I2πR2k^=−( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 6.0 cm )( 10 −2 m 1 cm ))k^+( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 6.0 cm )( 10 −2 m 1 cm ))k^=0 T

Conclusion:

Therefore, the magnetic field is 0 T .

(c)

To determine

The magnetic field at y=3.0 cm .

(c)

Expert Solution

Answer to Problem 25P

The magnetic field is 89 μT .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B=−μ0I2πR1k^+μ0I2πR2k^=−( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 9.0 cm )( 10 −2 m 1 cm ))k^+( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 3.0 cm )( 10 −2 m 1 cm ))k^=(( 89× 10 −6 T)( 10 6 μT 1 T ))k^=89 μT

Conclusion:

Therefore, the magnetic field is 89 μT .

(d)

To determine

The magnetic field at y=+9.0 cm .

(d)

Expert Solution

Answer to Problem 25P

The magnetic field is −160 μT .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B=−μ0I2πR1k^+μ0I2πR2k^=−( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 15.0 cm )( 10 −2 m 1 cm ))k^+( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 3.0 cm )( 10 −2 m 1 cm ))k^=−(( 160× 10 −6 T)( 10 6 μT 1 T ))k^=−160 μT

Conclusion:

Therefore, the magnetic field is −160 μT .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 27, Problem 25P

(a)

To determine

The magnetic field at y=−3.0 cm .

(a)

Expert Solution

Answer to Problem 25P

The magnetic field is −89 μT .

Explanation of Solution

Formula used:

The expression for magnetic field is given by,

B=−μ0I2πR1k^+μ0I2πR2k^

Calculation:

The magnetic field is calculated as,

B=−μ0I2πR1k^+μ0I2πR2k^=−( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 3.0 cm )( 10 −2 m 1 cm ))k^+( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 9.0 cm )( 10 −2 m 1 cm ))k^=−(( 89× 10 −6 T)( 10 6 μT 1 T ))k^=−89 μT

Conclusion:

Therefore, the magnetic fieldis −89 μT .

(b)

To determine

The magnetic field at y=0 .

(b)

Expert Solution

Answer to Problem 25P

The magnetic field is 0 T .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B=−μ0I2πR1k^+μ0I2πR2k^=−( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 6.0 cm )( 10 −2 m 1 cm ))k^+( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 6.0 cm )( 10 −2 m 1 cm ))k^=0 T

Conclusion:

Therefore, the magnetic field is 0 T .

(c)

To determine

The magnetic field at y=3.0 cm .

(c)

Expert Solution

Answer to Problem 25P

The magnetic field is 89 μT .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B=−μ0I2πR1k^+μ0I2πR2k^=−( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 9.0 cm )( 10 −2 m 1 cm ))k^+( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 3.0 cm )( 10 −2 m 1 cm ))k^=(( 89× 10 −6 T)( 10 6 μT 1 T ))k^=89 μT

Conclusion:

Therefore, the magnetic field is 89 μT .

(d)

To determine

The magnetic field at y=+9.0 cm .

(d)

Expert Solution

Answer to Problem 25P

The magnetic field is −160 μT .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B=−μ0I2πR1k^+μ0I2πR2k^=−( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 15.0 cm )( 10 −2 m 1 cm ))k^+( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 3.0 cm )( 10 −2 m 1 cm ))k^=−(( 160× 10 −6 T)( 10 6 μT 1 T ))k^=−160 μT

Conclusion:

Therefore, the magnetic field is −160 μT .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 27, Problem 25P

(a)

To determine

The magnetic field at y=−3.0 cm .

(a)

Expert Solution

Answer to Problem 25P

The magnetic field is −89 μT .

Explanation of Solution

Formula used:

The expression for magnetic field is given by,

B=−μ0I2πR1k^+μ0I2πR2k^

Calculation:

The magnetic field is calculated as,

B=−μ0I2πR1k^+μ0I2πR2k^=−( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 3.0 cm )( 10 −2 m 1 cm ))k^+( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 9.0 cm )( 10 −2 m 1 cm ))k^=−(( 89× 10 −6 T)( 10 6 μT 1 T ))k^=−89 μT

Conclusion:

Therefore, the magnetic fieldis −89 μT .

(b)

To determine

The magnetic field at y=0 .

(b)

Expert Solution

Answer to Problem 25P

The magnetic field is 0 T .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B=−μ0I2πR1k^+μ0I2πR2k^=−( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 6.0 cm )( 10 −2 m 1 cm ))k^+( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 6.0 cm )( 10 −2 m 1 cm ))k^=0 T

Conclusion:

Therefore, the magnetic field is 0 T .

(c)

To determine

The magnetic field at y=3.0 cm .

(c)

Expert Solution

Answer to Problem 25P

The magnetic field is 89 μT .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B=−μ0I2πR1k^+μ0I2πR2k^=−( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 9.0 cm )( 10 −2 m 1 cm ))k^+( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 3.0 cm )( 10 −2 m 1 cm ))k^=(( 89× 10 −6 T)( 10 6 μT 1 T ))k^=89 μT

Conclusion:

Therefore, the magnetic field is 89 μT .

(d)

To determine

The magnetic field at y=+9.0 cm .

(d)

Expert Solution

Answer to Problem 25P

The magnetic field is −160 μT .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B=−μ0I2πR1k^+μ0I2πR2k^=−( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 15.0 cm )( 10 −2 m 1 cm ))k^+( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 3.0 cm )( 10 −2 m 1 cm ))k^=−(( 160× 10 −6 T)( 10 6 μT 1 T ))k^=−160 μT

Conclusion:

Therefore, the magnetic field is −160 μT .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 27, Problem 25P

(a)

To determine

The magnetic field at y=−3.0 cm .

(a)

Expert Solution

Answer to Problem 25P

The magnetic field is −89 μT .

Explanation of Solution

Formula used:

The expression for magnetic field is given by,

B=−μ0I2πR1k^+μ0I2πR2k^

Calculation:

The magnetic field is calculated as,

B=−μ0I2πR1k^+μ0I2πR2k^=−( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 3.0 cm )( 10 −2 m 1 cm ))k^+( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 9.0 cm )( 10 −2 m 1 cm ))k^=−(( 89× 10 −6 T)( 10 6 μT 1 T ))k^=−89 μT

Conclusion:

Therefore, the magnetic fieldis −89 μT .

(b)

To determine

The magnetic field at y=0 .

(b)

Expert Solution

Answer to Problem 25P

The magnetic field is 0 T .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B=−μ0I2πR1k^+μ0I2πR2k^=−( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 6.0 cm )( 10 −2 m 1 cm ))k^+( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 6.0 cm )( 10 −2 m 1 cm ))k^=0 T

Conclusion:

Therefore, the magnetic field is 0 T .

(c)

To determine

The magnetic field at y=3.0 cm .

(c)

Expert Solution

Answer to Problem 25P

The magnetic field is 89 μT .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B=−μ0I2πR1k^+μ0I2πR2k^=−( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 9.0 cm )( 10 −2 m 1 cm ))k^+( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 3.0 cm )( 10 −2 m 1 cm ))k^=(( 89× 10 −6 T)( 10 6 μT 1 T ))k^=89 μT

Conclusion:

Therefore, the magnetic field is 89 μT .

(d)

To determine

The magnetic field at y=+9.0 cm .

(d)

Expert Solution

Answer to Problem 25P

The magnetic field is −160 μT .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B=−μ0I2πR1k^+μ0I2πR2k^=−( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 15.0 cm )( 10 −2 m 1 cm ))k^+( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 3.0 cm )( 10 −2 m 1 cm ))k^=−(( 160× 10 −6 T)( 10 6 μT 1 T ))k^=−160 μT

Conclusion:

Therefore, the magnetic field is −160 μT .

Answer 6

Chapter 27, Problem 25P

(a)

To determine

The magnetic field at y=−3.0 cm .

(a)

Expert Solution

Answer to Problem 25P

The magnetic field is −89 μT .

Explanation of Solution

Formula used:

The expression for magnetic field is given by,

B=−μ0I2πR1k^+μ0I2πR2k^

Calculation:

The magnetic field is calculated as,

B=−μ0I2πR1k^+μ0I2πR2k^=−( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 3.0 cm )( 10 −2 m 1 cm ))k^+( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 9.0 cm )( 10 −2 m 1 cm ))k^=−(( 89× 10 −6 T)( 10 6 μT 1 T ))k^=−89 μT

Conclusion:

Therefore, the magnetic fieldis −89 μT .

Answer 7

Chapter 27, Problem 25P

(a)

To determine

The magnetic field at y=−3.0 cm .

Answer 8

(a)

Expert Solution

Answer to Problem 25P

The magnetic field is −89 μT .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Formula used:

The expression for magnetic field is given by,

B=−μ0I2πR1k^+μ0I2πR2k^

Calculation:

The magnetic field is calculated as,

B=−μ0I2πR1k^+μ0I2πR2k^=−( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 3.0 cm )( 10 −2 m 1 cm ))k^+( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 9.0 cm )( 10 −2 m 1 cm ))k^=−(( 89× 10 −6 T)( 10 6 μT 1 T ))k^=−89 μT

Conclusion:

Therefore, the magnetic fieldis −89 μT .

Answer 13

(b)

To determine

The magnetic field at y=0 .

(b)

Expert Solution

Answer to Problem 25P

The magnetic field is 0 T .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B=−μ0I2πR1k^+μ0I2πR2k^=−( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 6.0 cm )( 10 −2 m 1 cm ))k^+( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 6.0 cm )( 10 −2 m 1 cm ))k^=0 T

Conclusion:

Therefore, the magnetic field is 0 T .

Answer 14

(b)

To determine

The magnetic field at y=0 .

Answer 15

(b)

Expert Solution

Answer to Problem 25P

The magnetic field is 0 T .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B=−μ0I2πR1k^+μ0I2πR2k^=−( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 6.0 cm )( 10 −2 m 1 cm ))k^+( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 6.0 cm )( 10 −2 m 1 cm ))k^=0 T

Conclusion:

Therefore, the magnetic field is 0 T .

Answer 20

(c)

To determine

The magnetic field at y=3.0 cm .

(c)

Expert Solution

Answer to Problem 25P

The magnetic field is 89 μT .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B=−μ0I2πR1k^+μ0I2πR2k^=−( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 9.0 cm )( 10 −2 m 1 cm ))k^+( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 3.0 cm )( 10 −2 m 1 cm ))k^=(( 89× 10 −6 T)( 10 6 μT 1 T ))k^=89 μT

Conclusion:

Therefore, the magnetic field is 89 μT .

Answer 21

(c)

To determine

The magnetic field at y=3.0 cm .

Answer 22

(c)

Expert Solution

Answer to Problem 25P

The magnetic field is 89 μT .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B=−μ0I2πR1k^+μ0I2πR2k^=−( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 9.0 cm )( 10 −2 m 1 cm ))k^+( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 3.0 cm )( 10 −2 m 1 cm ))k^=(( 89× 10 −6 T)( 10 6 μT 1 T ))k^=89 μT

Conclusion:

Therefore, the magnetic field is 89 μT .

Answer 27

(d)

To determine

The magnetic field at y=+9.0 cm .

(d)

Expert Solution

Answer to Problem 25P

The magnetic field is −160 μT .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B=−μ0I2πR1k^+μ0I2πR2k^=−( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 15.0 cm )( 10 −2 m 1 cm ))k^+( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 3.0 cm )( 10 −2 m 1 cm ))k^=−(( 160× 10 −6 T)( 10 6 μT 1 T ))k^=−160 μT

Conclusion:

Therefore, the magnetic field is −160 μT .

Answer 28

(d)

To determine

The magnetic field at y=+9.0 cm .

Answer 29

(d)

Expert Solution

Answer to Problem 25P

The magnetic field is −160 μT .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B=−μ0I2πR1k^+μ0I2πR2k^=−( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 15.0 cm )( 10 −2 m 1 cm ))k^+( 4π× 10 −7 T⋅m/A )( 20 A)2π( ( 3.0 cm )( 10 −2 m 1 cm ))k^=−(( 160× 10 −6 T)( 10 6 μT 1 T ))k^=−160 μT