The magnetic field at x = 1.0 m , y = 3.0 m .

Question

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Answer 1

Question

Chapter 27, Problem 15P

(a)

To determine

The magnetic field at x=1.0 m,y=3.0 m .

(a)

Expert Solution

Answer to Problem 15P

The magnetic field is 0 .

Explanation of Solution

Given:

The velocity is v→=1.0×102 m/s i^+2.0×102 m/s j^ .

Formula used:

The expression for magnetic field is given by,

B→=μ04πqv→×r^r2

Calculation:

The magnetic field is calculated as,

B → = μ 0 4π q v → × r ^ r 2

=( 4π× 10 −7 T⋅m/A 4π )( ( ( 1.602× 10 −19 C ) )( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )× r ^ ) ( ( 2.0 m−3.0 m ) i ^ +( 2.0 m−4.0 m ) j ^ ) 2 )

=[ ( ( 36× 10 −12 T⋅ m 2 )( 10 12 pT 1 T ) ) ( ( i ^ ×( − 1 ( −1.0 m ) 2 + ( −2.0 m ) 2 i ^ − 2 ( −1.0 m ) 2 + ( −2.0 m ) 2 j ^ ) ) ( ( −1.0 m ) 2 + ( −2.0 m ) 2 ) 2 ) ]

=0

Conclusion:

Therefore, the magnetic field at origin is 0 .

(b)

To determine

The magnetic field at x=6.0, y=4.0 m .

(b)

Expert Solution

Answer to Problem 15P

The magnetic field is (−3.6×10−23 T)k^ .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B→=μ04πqv→×r^r2=( 4π× 10 −7 T⋅m/A 4π)( ( ( 1.602× 10 −19 C ) )( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )× r ^ ) ( ( 6.0 m−3.0 m ) i ^ −( 4.0 m−4.0 m ) j ^ ) 2 )=(( 36× 10 −12 T⋅ m 2 )( 10 12 pT 1 T ))( ( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )×r( ( 3.0 m ) i ^ 3.0 m ) ) ( 3.0 ) 2 )=(−3.6× 10 −23 T)k^

Conclusion:

Therefore, the magnetic field is (−3.6×10−23 T)k^ .

(c)

To determine

The magnetic field at x=3.0, y=6.0 m .

(c)

Expert Solution

Answer to Problem 15P

The magnetic field is (4.6×10−23 T)k^ .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B→=μ04πqv→×r^r2=( 4π× 10 −7 T⋅m/A 4π)( ( ( 1.602× 10 −19 C ) )( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )× r ^ ) ( ( 3.0 m−3.0 m ) i ^ −( 6.0 m−4.0 m ) j ^ ) 2 )=(( 36× 10 −12 T⋅ m 2 )( 10 12 pT 1 T ))( ( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )×r( ( 2.0 m ) i ^ 2.0 m ) ) ( 2.0 ) 2 )=(4.6× 10 −23 T)k^

Conclusion:

Therefore, the magnetic field is (4.6×10−23 T)k^ .

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Answer 2

Question

Chapter 27, Problem 15P

(a)

To determine

The magnetic field at x=1.0 m,y=3.0 m .

(a)

Expert Solution

Answer to Problem 15P

The magnetic field is 0 .

Explanation of Solution

Given:

The velocity is v→=1.0×102 m/s i^+2.0×102 m/s j^ .

Formula used:

The expression for magnetic field is given by,

B→=μ04πqv→×r^r2

Calculation:

The magnetic field is calculated as,

B → = μ 0 4π q v → × r ^ r 2

=( 4π× 10 −7 T⋅m/A 4π )( ( ( 1.602× 10 −19 C ) )( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )× r ^ ) ( ( 2.0 m−3.0 m ) i ^ +( 2.0 m−4.0 m ) j ^ ) 2 )

=[ ( ( 36× 10 −12 T⋅ m 2 )( 10 12 pT 1 T ) ) ( ( i ^ ×( − 1 ( −1.0 m ) 2 + ( −2.0 m ) 2 i ^ − 2 ( −1.0 m ) 2 + ( −2.0 m ) 2 j ^ ) ) ( ( −1.0 m ) 2 + ( −2.0 m ) 2 ) 2 ) ]

=0

Conclusion:

Therefore, the magnetic field at origin is 0 .

(b)

To determine

The magnetic field at x=6.0, y=4.0 m .

(b)

Expert Solution

Answer to Problem 15P

The magnetic field is (−3.6×10−23 T)k^ .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B→=μ04πqv→×r^r2=( 4π× 10 −7 T⋅m/A 4π)( ( ( 1.602× 10 −19 C ) )( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )× r ^ ) ( ( 6.0 m−3.0 m ) i ^ −( 4.0 m−4.0 m ) j ^ ) 2 )=(( 36× 10 −12 T⋅ m 2 )( 10 12 pT 1 T ))( ( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )×r( ( 3.0 m ) i ^ 3.0 m ) ) ( 3.0 ) 2 )=(−3.6× 10 −23 T)k^

Conclusion:

Therefore, the magnetic field is (−3.6×10−23 T)k^ .

(c)

To determine

The magnetic field at x=3.0, y=6.0 m .

(c)

Expert Solution

Answer to Problem 15P

The magnetic field is (4.6×10−23 T)k^ .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B→=μ04πqv→×r^r2=( 4π× 10 −7 T⋅m/A 4π)( ( ( 1.602× 10 −19 C ) )( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )× r ^ ) ( ( 3.0 m−3.0 m ) i ^ −( 6.0 m−4.0 m ) j ^ ) 2 )=(( 36× 10 −12 T⋅ m 2 )( 10 12 pT 1 T ))( ( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )×r( ( 2.0 m ) i ^ 2.0 m ) ) ( 2.0 ) 2 )=(4.6× 10 −23 T)k^

Conclusion:

Therefore, the magnetic field is (4.6×10−23 T)k^ .

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Answer 3

Question

Chapter 27, Problem 15P

(a)

To determine

The magnetic field at x=1.0 m,y=3.0 m .

(a)

Expert Solution

Answer to Problem 15P

The magnetic field is 0 .

Explanation of Solution

Given:

The velocity is v→=1.0×102 m/s i^+2.0×102 m/s j^ .

Formula used:

The expression for magnetic field is given by,

B→=μ04πqv→×r^r2

Calculation:

The magnetic field is calculated as,

B → = μ 0 4π q v → × r ^ r 2

=( 4π× 10 −7 T⋅m/A 4π )( ( ( 1.602× 10 −19 C ) )( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )× r ^ ) ( ( 2.0 m−3.0 m ) i ^ +( 2.0 m−4.0 m ) j ^ ) 2 )

=[ ( ( 36× 10 −12 T⋅ m 2 )( 10 12 pT 1 T ) ) ( ( i ^ ×( − 1 ( −1.0 m ) 2 + ( −2.0 m ) 2 i ^ − 2 ( −1.0 m ) 2 + ( −2.0 m ) 2 j ^ ) ) ( ( −1.0 m ) 2 + ( −2.0 m ) 2 ) 2 ) ]

=0

Conclusion:

Therefore, the magnetic field at origin is 0 .

(b)

To determine

The magnetic field at x=6.0, y=4.0 m .

(b)

Expert Solution

Answer to Problem 15P

The magnetic field is (−3.6×10−23 T)k^ .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B→=μ04πqv→×r^r2=( 4π× 10 −7 T⋅m/A 4π)( ( ( 1.602× 10 −19 C ) )( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )× r ^ ) ( ( 6.0 m−3.0 m ) i ^ −( 4.0 m−4.0 m ) j ^ ) 2 )=(( 36× 10 −12 T⋅ m 2 )( 10 12 pT 1 T ))( ( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )×r( ( 3.0 m ) i ^ 3.0 m ) ) ( 3.0 ) 2 )=(−3.6× 10 −23 T)k^

Conclusion:

Therefore, the magnetic field is (−3.6×10−23 T)k^ .

(c)

To determine

The magnetic field at x=3.0, y=6.0 m .

(c)

Expert Solution

Answer to Problem 15P

The magnetic field is (4.6×10−23 T)k^ .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B→=μ04πqv→×r^r2=( 4π× 10 −7 T⋅m/A 4π)( ( ( 1.602× 10 −19 C ) )( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )× r ^ ) ( ( 3.0 m−3.0 m ) i ^ −( 6.0 m−4.0 m ) j ^ ) 2 )=(( 36× 10 −12 T⋅ m 2 )( 10 12 pT 1 T ))( ( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )×r( ( 2.0 m ) i ^ 2.0 m ) ) ( 2.0 ) 2 )=(4.6× 10 −23 T)k^

Conclusion:

Therefore, the magnetic field is (4.6×10−23 T)k^ .

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Answer 4

Question

Chapter 27, Problem 15P

(a)

To determine

The magnetic field at x=1.0 m,y=3.0 m .

(a)

Expert Solution

Answer to Problem 15P

The magnetic field is 0 .

Explanation of Solution

Given:

The velocity is v→=1.0×102 m/s i^+2.0×102 m/s j^ .

Formula used:

The expression for magnetic field is given by,

B→=μ04πqv→×r^r2

Calculation:

The magnetic field is calculated as,

B → = μ 0 4π q v → × r ^ r 2

=( 4π× 10 −7 T⋅m/A 4π )( ( ( 1.602× 10 −19 C ) )( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )× r ^ ) ( ( 2.0 m−3.0 m ) i ^ +( 2.0 m−4.0 m ) j ^ ) 2 )

=[ ( ( 36× 10 −12 T⋅ m 2 )( 10 12 pT 1 T ) ) ( ( i ^ ×( − 1 ( −1.0 m ) 2 + ( −2.0 m ) 2 i ^ − 2 ( −1.0 m ) 2 + ( −2.0 m ) 2 j ^ ) ) ( ( −1.0 m ) 2 + ( −2.0 m ) 2 ) 2 ) ]

=0

Conclusion:

Therefore, the magnetic field at origin is 0 .

(b)

To determine

The magnetic field at x=6.0, y=4.0 m .

(b)

Expert Solution

Answer to Problem 15P

The magnetic field is (−3.6×10−23 T)k^ .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B→=μ04πqv→×r^r2=( 4π× 10 −7 T⋅m/A 4π)( ( ( 1.602× 10 −19 C ) )( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )× r ^ ) ( ( 6.0 m−3.0 m ) i ^ −( 4.0 m−4.0 m ) j ^ ) 2 )=(( 36× 10 −12 T⋅ m 2 )( 10 12 pT 1 T ))( ( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )×r( ( 3.0 m ) i ^ 3.0 m ) ) ( 3.0 ) 2 )=(−3.6× 10 −23 T)k^

Conclusion:

Therefore, the magnetic field is (−3.6×10−23 T)k^ .

(c)

To determine

The magnetic field at x=3.0, y=6.0 m .

(c)

Expert Solution

Answer to Problem 15P

The magnetic field is (4.6×10−23 T)k^ .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B→=μ04πqv→×r^r2=( 4π× 10 −7 T⋅m/A 4π)( ( ( 1.602× 10 −19 C ) )( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )× r ^ ) ( ( 3.0 m−3.0 m ) i ^ −( 6.0 m−4.0 m ) j ^ ) 2 )=(( 36× 10 −12 T⋅ m 2 )( 10 12 pT 1 T ))( ( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )×r( ( 2.0 m ) i ^ 2.0 m ) ) ( 2.0 ) 2 )=(4.6× 10 −23 T)k^

Conclusion:

Therefore, the magnetic field is (4.6×10−23 T)k^ .

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Answer 5

Chapter 27, Problem 15P

(a)

To determine

The magnetic field at x=1.0 m,y=3.0 m .

(a)

Expert Solution

Answer to Problem 15P

The magnetic field is 0 .

Explanation of Solution

Given:

The velocity is v→=1.0×102 m/s i^+2.0×102 m/s j^ .

Formula used:

The expression for magnetic field is given by,

B→=μ04πqv→×r^r2

Calculation:

The magnetic field is calculated as,

B → = μ 0 4π q v → × r ^ r 2

=( 4π× 10 −7 T⋅m/A 4π )( ( ( 1.602× 10 −19 C ) )( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )× r ^ ) ( ( 2.0 m−3.0 m ) i ^ +( 2.0 m−4.0 m ) j ^ ) 2 )

=[ ( ( 36× 10 −12 T⋅ m 2 )( 10 12 pT 1 T ) ) ( ( i ^ ×( − 1 ( −1.0 m ) 2 + ( −2.0 m ) 2 i ^ − 2 ( −1.0 m ) 2 + ( −2.0 m ) 2 j ^ ) ) ( ( −1.0 m ) 2 + ( −2.0 m ) 2 ) 2 ) ]

=0

Conclusion:

Therefore, the magnetic field at origin is 0 .

(b)

To determine

The magnetic field at x=6.0, y=4.0 m .

(b)

Expert Solution

Answer to Problem 15P

The magnetic field is (−3.6×10−23 T)k^ .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B→=μ04πqv→×r^r2=( 4π× 10 −7 T⋅m/A 4π)( ( ( 1.602× 10 −19 C ) )( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )× r ^ ) ( ( 6.0 m−3.0 m ) i ^ −( 4.0 m−4.0 m ) j ^ ) 2 )=(( 36× 10 −12 T⋅ m 2 )( 10 12 pT 1 T ))( ( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )×r( ( 3.0 m ) i ^ 3.0 m ) ) ( 3.0 ) 2 )=(−3.6× 10 −23 T)k^

Conclusion:

Therefore, the magnetic field is (−3.6×10−23 T)k^ .

(c)

To determine

The magnetic field at x=3.0, y=6.0 m .

(c)

Expert Solution

Answer to Problem 15P

The magnetic field is (4.6×10−23 T)k^ .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B→=μ04πqv→×r^r2=( 4π× 10 −7 T⋅m/A 4π)( ( ( 1.602× 10 −19 C ) )( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )× r ^ ) ( ( 3.0 m−3.0 m ) i ^ −( 6.0 m−4.0 m ) j ^ ) 2 )=(( 36× 10 −12 T⋅ m 2 )( 10 12 pT 1 T ))( ( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )×r( ( 2.0 m ) i ^ 2.0 m ) ) ( 2.0 ) 2 )=(4.6× 10 −23 T)k^

Conclusion:

Therefore, the magnetic field is (4.6×10−23 T)k^ .

Answer 6

Chapter 27, Problem 15P

(a)

To determine

The magnetic field at x=1.0 m,y=3.0 m .

(a)

Expert Solution

Answer to Problem 15P

The magnetic field is 0 .

Explanation of Solution

Given:

The velocity is v→=1.0×102 m/s i^+2.0×102 m/s j^ .

Formula used:

The expression for magnetic field is given by,

B→=μ04πqv→×r^r2

Calculation:

The magnetic field is calculated as,

B → = μ 0 4π q v → × r ^ r 2

=( 4π× 10 −7 T⋅m/A 4π )( ( ( 1.602× 10 −19 C ) )( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )× r ^ ) ( ( 2.0 m−3.0 m ) i ^ +( 2.0 m−4.0 m ) j ^ ) 2 )

=[ ( ( 36× 10 −12 T⋅ m 2 )( 10 12 pT 1 T ) ) ( ( i ^ ×( − 1 ( −1.0 m ) 2 + ( −2.0 m ) 2 i ^ − 2 ( −1.0 m ) 2 + ( −2.0 m ) 2 j ^ ) ) ( ( −1.0 m ) 2 + ( −2.0 m ) 2 ) 2 ) ]

=0

Conclusion:

Therefore, the magnetic field at origin is 0 .

Answer 7

Chapter 27, Problem 15P

(a)

To determine

The magnetic field at x=1.0 m,y=3.0 m .

Answer 8

(a)

Expert Solution

Answer to Problem 15P

The magnetic field is 0 .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

The velocity is v→=1.0×102 m/s i^+2.0×102 m/s j^ .

Formula used:

The expression for magnetic field is given by,

B→=μ04πqv→×r^r2

Calculation:

The magnetic field is calculated as,

B → = μ 0 4π q v → × r ^ r 2

=( 4π× 10 −7 T⋅m/A 4π )( ( ( 1.602× 10 −19 C ) )( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )× r ^ ) ( ( 2.0 m−3.0 m ) i ^ +( 2.0 m−4.0 m ) j ^ ) 2 )

=[ ( ( 36× 10 −12 T⋅ m 2 )( 10 12 pT 1 T ) ) ( ( i ^ ×( − 1 ( −1.0 m ) 2 + ( −2.0 m ) 2 i ^ − 2 ( −1.0 m ) 2 + ( −2.0 m ) 2 j ^ ) ) ( ( −1.0 m ) 2 + ( −2.0 m ) 2 ) 2 ) ]

=0

Conclusion:

Therefore, the magnetic field at origin is 0 .

Answer 13

(b)

To determine

The magnetic field at x=6.0, y=4.0 m .

(b)

Expert Solution

Answer to Problem 15P

The magnetic field is (−3.6×10−23 T)k^ .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B→=μ04πqv→×r^r2=( 4π× 10 −7 T⋅m/A 4π)( ( ( 1.602× 10 −19 C ) )( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )× r ^ ) ( ( 6.0 m−3.0 m ) i ^ −( 4.0 m−4.0 m ) j ^ ) 2 )=(( 36× 10 −12 T⋅ m 2 )( 10 12 pT 1 T ))( ( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )×r( ( 3.0 m ) i ^ 3.0 m ) ) ( 3.0 ) 2 )=(−3.6× 10 −23 T)k^

Conclusion:

Therefore, the magnetic field is (−3.6×10−23 T)k^ .

Answer 14

(b)

To determine

The magnetic field at x=6.0, y=4.0 m .

Answer 15

(b)

Expert Solution

Answer to Problem 15P

The magnetic field is (−3.6×10−23 T)k^ .

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→=μ04πqv→×r^r2=( 4π× 10 −7 T⋅m/A 4π)( ( ( 1.602× 10 −19 C ) )( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )× r ^ ) ( ( 6.0 m−3.0 m ) i ^ −( 4.0 m−4.0 m ) j ^ ) 2 )=(( 36× 10 −12 T⋅ m 2 )( 10 12 pT 1 T ))( ( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )×r( ( 3.0 m ) i ^ 3.0 m ) ) ( 3.0 ) 2 )=(−3.6× 10 −23 T)k^

Conclusion:

Therefore, the magnetic field is (−3.6×10−23 T)k^ .

Answer 20

(c)

To determine

The magnetic field at x=3.0, y=6.0 m .

(c)

Expert Solution

Answer to Problem 15P

The magnetic field is (4.6×10−23 T)k^ .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

B→=μ04πqv→×r^r2=( 4π× 10 −7 T⋅m/A 4π)( ( ( 1.602× 10 −19 C ) )( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )× r ^ ) ( ( 3.0 m−3.0 m ) i ^ −( 6.0 m−4.0 m ) j ^ ) 2 )=(( 36× 10 −12 T⋅ m 2 )( 10 12 pT 1 T ))( ( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )×r( ( 2.0 m ) i ^ 2.0 m ) ) ( 2.0 ) 2 )=(4.6× 10 −23 T)k^

Conclusion:

Therefore, the magnetic field is (4.6×10−23 T)k^ .

Answer 21

(c)

To determine

The magnetic field at x=3.0, y=6.0 m .

Answer 22

(c)

Expert Solution

Answer to Problem 15P

The magnetic field is (4.6×10−23 T)k^ .

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→=μ04πqv→×r^r2=( 4π× 10 −7 T⋅m/A 4π)( ( ( 1.602× 10 −19 C ) )( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )× r ^ ) ( ( 3.0 m−3.0 m ) i ^ −( 6.0 m−4.0 m ) j ^ ) 2 )=(( 36× 10 −12 T⋅ m 2 )( 10 12 pT 1 T ))( ( ( 1.0× 10 2 m/s i ^ +2.0× 10 2 m/s j ^ )×r( ( 2.0 m ) i ^ 2.0 m ) ) ( 2.0 ) 2 )=(4.6× 10 −23 T)k^