The force on each segment of rectangular coil.

Question

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Answer 1

Question

Chapter 27, Problem 81P

(a)

To determine

The force on each segment of rectangular coil.

(a)

Expert Solution

Answer to Problem 81P

The force on segment 1, 2, 3 and 4 are −(2.5×10−5 N)(j^) , (1.00×10−4 N)(i^) , (2.5×10−5 N)(j^) and (−0.286×10−4 N)(i^) respectively.

Explanation of Solution

Given:

The straight wire carries a current I1=20.0 A .

The coil carries a current I2=5.00 A .

The length of parallel sides are 5.00 cm and 10.0 cm .

Formula used:

The expression for force in first segment is given by,

F1=−μ0I1I22πlog(7.0 cm2.0 cm)(j^)

The expression for force in second segment is given by,

F2=−μ0I1I2l22π(2.0 cm)(i^)

The expression for force in third segment is given by,

F3=−F1

The expression for force in fourth segment is given by,

F4=−μ0I1I2l42π(7.0 cm)(i^)

Calculation:

The force in first segment is calculated as,

F1=−μ0I1I22πlog( 7.0 cm 2.0 cm)(j^)=−( 4π× 10 −7 N/ A 2 )( 20.0 A)( 5.00 A)2πlog( 7.0 cm 2.0 cm)(j^)=−(2.5× 10 −5 N)(j^)

The force in second segment is calculated as,

F2=−μ0I1I2l22π( 2.0 cm)(i^)=−( 4π× 10 −7 N/ A 2 )( 20.0 A)( 5.00 A)( ( 10 cm )( 10 2 m 1 cm ))2π( ( 2.0 cm )( 10 2 m 1 cm ))l(i^)=(1.00× 10 −4 N)(i^)

The expression for force in third segment is given by,

F3=−F1=−(−2.5× 10 −5 N)(j^)=(2.5× 10 −5 N)(j^)

The expression for force in fourth segment is given by,

F4=−μ0I1I2l42π( 7.0 cm)(i^)=−( 4π× 10 −7 N/ A 2 )( 20.0 A)( 5.00 A)( ( 10 cm )( 10 2 m 1 cm ))2( ( 7.0 cm )( 10 2 m 1 cm ))(i^)=(−0.286× 10 −4 N)(i^)

Conclusion:

Therefore, the force on segment 1, 2, 3 and 4 are −(2.5×10−5 N)(j^) , (1.00×10−4 N)(i^) , (2.5×10−5 N)(j^) and (−0.286×10−4 N)(i^) respectively.

(b)

To determine

The net force on the coil

(b)

Expert Solution

Answer to Problem 81P

The net force on the coil is (0.71×10−4 N)i^ .

Explanation of Solution

Formula used:

The expression for net force on the coil is given by,

Fnet=F1+F2+F3+F4

Calculation:

The net force on the coil is calculated as,

Fnet=F1+F2+F3+F4=−(2.5× 10 −5 N)(j^)+(( 1.00× 10 −4 N)( i ^))+(( 2.5× 10 −5 N)( j ^))+(−0.286× 10 −4 N)(i^)=(1.00× 10 −4 N)i^+(−0.286× 10 −4 N)(i^)=(0.71× 10 −4 N)i^

Conclusion:

Therefore, the net force on the coil is (0.71×10−4 N)i^ .

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Answer 2

Question

Chapter 27, Problem 81P

(a)

To determine

The force on each segment of rectangular coil.

(a)

Expert Solution

Answer to Problem 81P

The force on segment 1, 2, 3 and 4 are −(2.5×10−5 N)(j^) , (1.00×10−4 N)(i^) , (2.5×10−5 N)(j^) and (−0.286×10−4 N)(i^) respectively.

Explanation of Solution

Given:

The straight wire carries a current I1=20.0 A .

The coil carries a current I2=5.00 A .

The length of parallel sides are 5.00 cm and 10.0 cm .

Formula used:

The expression for force in first segment is given by,

F1=−μ0I1I22πlog(7.0 cm2.0 cm)(j^)

The expression for force in second segment is given by,

F2=−μ0I1I2l22π(2.0 cm)(i^)

The expression for force in third segment is given by,

F3=−F1

The expression for force in fourth segment is given by,

F4=−μ0I1I2l42π(7.0 cm)(i^)

Calculation:

The force in first segment is calculated as,

F1=−μ0I1I22πlog( 7.0 cm 2.0 cm)(j^)=−( 4π× 10 −7 N/ A 2 )( 20.0 A)( 5.00 A)2πlog( 7.0 cm 2.0 cm)(j^)=−(2.5× 10 −5 N)(j^)

The force in second segment is calculated as,

F2=−μ0I1I2l22π( 2.0 cm)(i^)=−( 4π× 10 −7 N/ A 2 )( 20.0 A)( 5.00 A)( ( 10 cm )( 10 2 m 1 cm ))2π( ( 2.0 cm )( 10 2 m 1 cm ))l(i^)=(1.00× 10 −4 N)(i^)

The expression for force in third segment is given by,

F3=−F1=−(−2.5× 10 −5 N)(j^)=(2.5× 10 −5 N)(j^)

The expression for force in fourth segment is given by,

F4=−μ0I1I2l42π( 7.0 cm)(i^)=−( 4π× 10 −7 N/ A 2 )( 20.0 A)( 5.00 A)( ( 10 cm )( 10 2 m 1 cm ))2( ( 7.0 cm )( 10 2 m 1 cm ))(i^)=(−0.286× 10 −4 N)(i^)

Conclusion:

Therefore, the force on segment 1, 2, 3 and 4 are −(2.5×10−5 N)(j^) , (1.00×10−4 N)(i^) , (2.5×10−5 N)(j^) and (−0.286×10−4 N)(i^) respectively.

(b)

To determine

The net force on the coil

(b)

Expert Solution

Answer to Problem 81P

The net force on the coil is (0.71×10−4 N)i^ .

Explanation of Solution

Formula used:

The expression for net force on the coil is given by,

Fnet=F1+F2+F3+F4

Calculation:

The net force on the coil is calculated as,

Fnet=F1+F2+F3+F4=−(2.5× 10 −5 N)(j^)+(( 1.00× 10 −4 N)( i ^))+(( 2.5× 10 −5 N)( j ^))+(−0.286× 10 −4 N)(i^)=(1.00× 10 −4 N)i^+(−0.286× 10 −4 N)(i^)=(0.71× 10 −4 N)i^

Conclusion:

Therefore, the net force on the coil is (0.71×10−4 N)i^ .

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Answer 3

Question

Chapter 27, Problem 81P

(a)

To determine

The force on each segment of rectangular coil.

(a)

Expert Solution

Answer to Problem 81P

The force on segment 1, 2, 3 and 4 are −(2.5×10−5 N)(j^) , (1.00×10−4 N)(i^) , (2.5×10−5 N)(j^) and (−0.286×10−4 N)(i^) respectively.

Explanation of Solution

Given:

The straight wire carries a current I1=20.0 A .

The coil carries a current I2=5.00 A .

The length of parallel sides are 5.00 cm and 10.0 cm .

Formula used:

The expression for force in first segment is given by,

F1=−μ0I1I22πlog(7.0 cm2.0 cm)(j^)

The expression for force in second segment is given by,

F2=−μ0I1I2l22π(2.0 cm)(i^)

The expression for force in third segment is given by,

F3=−F1

The expression for force in fourth segment is given by,

F4=−μ0I1I2l42π(7.0 cm)(i^)

Calculation:

The force in first segment is calculated as,

F1=−μ0I1I22πlog( 7.0 cm 2.0 cm)(j^)=−( 4π× 10 −7 N/ A 2 )( 20.0 A)( 5.00 A)2πlog( 7.0 cm 2.0 cm)(j^)=−(2.5× 10 −5 N)(j^)

The force in second segment is calculated as,

F2=−μ0I1I2l22π( 2.0 cm)(i^)=−( 4π× 10 −7 N/ A 2 )( 20.0 A)( 5.00 A)( ( 10 cm )( 10 2 m 1 cm ))2π( ( 2.0 cm )( 10 2 m 1 cm ))l(i^)=(1.00× 10 −4 N)(i^)

The expression for force in third segment is given by,

F3=−F1=−(−2.5× 10 −5 N)(j^)=(2.5× 10 −5 N)(j^)

The expression for force in fourth segment is given by,

F4=−μ0I1I2l42π( 7.0 cm)(i^)=−( 4π× 10 −7 N/ A 2 )( 20.0 A)( 5.00 A)( ( 10 cm )( 10 2 m 1 cm ))2( ( 7.0 cm )( 10 2 m 1 cm ))(i^)=(−0.286× 10 −4 N)(i^)

Conclusion:

Therefore, the force on segment 1, 2, 3 and 4 are −(2.5×10−5 N)(j^) , (1.00×10−4 N)(i^) , (2.5×10−5 N)(j^) and (−0.286×10−4 N)(i^) respectively.

(b)

To determine

The net force on the coil

(b)

Expert Solution

Answer to Problem 81P

The net force on the coil is (0.71×10−4 N)i^ .

Explanation of Solution

Formula used:

The expression for net force on the coil is given by,

Fnet=F1+F2+F3+F4

Calculation:

The net force on the coil is calculated as,

Fnet=F1+F2+F3+F4=−(2.5× 10 −5 N)(j^)+(( 1.00× 10 −4 N)( i ^))+(( 2.5× 10 −5 N)( j ^))+(−0.286× 10 −4 N)(i^)=(1.00× 10 −4 N)i^+(−0.286× 10 −4 N)(i^)=(0.71× 10 −4 N)i^

Conclusion:

Therefore, the net force on the coil is (0.71×10−4 N)i^ .

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Answer 4

Question

Chapter 27, Problem 81P

(a)

To determine

The force on each segment of rectangular coil.

(a)

Expert Solution

Answer to Problem 81P

The force on segment 1, 2, 3 and 4 are −(2.5×10−5 N)(j^) , (1.00×10−4 N)(i^) , (2.5×10−5 N)(j^) and (−0.286×10−4 N)(i^) respectively.

Explanation of Solution

Given:

The straight wire carries a current I1=20.0 A .

The coil carries a current I2=5.00 A .

The length of parallel sides are 5.00 cm and 10.0 cm .

Formula used:

The expression for force in first segment is given by,

F1=−μ0I1I22πlog(7.0 cm2.0 cm)(j^)

The expression for force in second segment is given by,

F2=−μ0I1I2l22π(2.0 cm)(i^)

The expression for force in third segment is given by,

F3=−F1

The expression for force in fourth segment is given by,

F4=−μ0I1I2l42π(7.0 cm)(i^)

Calculation:

The force in first segment is calculated as,

F1=−μ0I1I22πlog( 7.0 cm 2.0 cm)(j^)=−( 4π× 10 −7 N/ A 2 )( 20.0 A)( 5.00 A)2πlog( 7.0 cm 2.0 cm)(j^)=−(2.5× 10 −5 N)(j^)

The force in second segment is calculated as,

F2=−μ0I1I2l22π( 2.0 cm)(i^)=−( 4π× 10 −7 N/ A 2 )( 20.0 A)( 5.00 A)( ( 10 cm )( 10 2 m 1 cm ))2π( ( 2.0 cm )( 10 2 m 1 cm ))l(i^)=(1.00× 10 −4 N)(i^)

The expression for force in third segment is given by,

F3=−F1=−(−2.5× 10 −5 N)(j^)=(2.5× 10 −5 N)(j^)

The expression for force in fourth segment is given by,

F4=−μ0I1I2l42π( 7.0 cm)(i^)=−( 4π× 10 −7 N/ A 2 )( 20.0 A)( 5.00 A)( ( 10 cm )( 10 2 m 1 cm ))2( ( 7.0 cm )( 10 2 m 1 cm ))(i^)=(−0.286× 10 −4 N)(i^)

Conclusion:

Therefore, the force on segment 1, 2, 3 and 4 are −(2.5×10−5 N)(j^) , (1.00×10−4 N)(i^) , (2.5×10−5 N)(j^) and (−0.286×10−4 N)(i^) respectively.

(b)

To determine

The net force on the coil

(b)

Expert Solution

Answer to Problem 81P

The net force on the coil is (0.71×10−4 N)i^ .

Explanation of Solution

Formula used:

The expression for net force on the coil is given by,

Fnet=F1+F2+F3+F4

Calculation:

The net force on the coil is calculated as,

Fnet=F1+F2+F3+F4=−(2.5× 10 −5 N)(j^)+(( 1.00× 10 −4 N)( i ^))+(( 2.5× 10 −5 N)( j ^))+(−0.286× 10 −4 N)(i^)=(1.00× 10 −4 N)i^+(−0.286× 10 −4 N)(i^)=(0.71× 10 −4 N)i^

Conclusion:

Therefore, the net force on the coil is (0.71×10−4 N)i^ .

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Answer 5

Chapter 27, Problem 81P

(a)

To determine

The force on each segment of rectangular coil.

(a)

Expert Solution

Answer to Problem 81P

The force on segment 1, 2, 3 and 4 are −(2.5×10−5 N)(j^) , (1.00×10−4 N)(i^) , (2.5×10−5 N)(j^) and (−0.286×10−4 N)(i^) respectively.

Explanation of Solution

Given:

The straight wire carries a current I1=20.0 A .

The coil carries a current I2=5.00 A .

The length of parallel sides are 5.00 cm and 10.0 cm .

Formula used:

The expression for force in first segment is given by,

F1=−μ0I1I22πlog(7.0 cm2.0 cm)(j^)

The expression for force in second segment is given by,

F2=−μ0I1I2l22π(2.0 cm)(i^)

The expression for force in third segment is given by,

F3=−F1

The expression for force in fourth segment is given by,

F4=−μ0I1I2l42π(7.0 cm)(i^)

Calculation:

The force in first segment is calculated as,

F1=−μ0I1I22πlog( 7.0 cm 2.0 cm)(j^)=−( 4π× 10 −7 N/ A 2 )( 20.0 A)( 5.00 A)2πlog( 7.0 cm 2.0 cm)(j^)=−(2.5× 10 −5 N)(j^)

The force in second segment is calculated as,

F2=−μ0I1I2l22π( 2.0 cm)(i^)=−( 4π× 10 −7 N/ A 2 )( 20.0 A)( 5.00 A)( ( 10 cm )( 10 2 m 1 cm ))2π( ( 2.0 cm )( 10 2 m 1 cm ))l(i^)=(1.00× 10 −4 N)(i^)

The expression for force in third segment is given by,

F3=−F1=−(−2.5× 10 −5 N)(j^)=(2.5× 10 −5 N)(j^)

The expression for force in fourth segment is given by,

F4=−μ0I1I2l42π( 7.0 cm)(i^)=−( 4π× 10 −7 N/ A 2 )( 20.0 A)( 5.00 A)( ( 10 cm )( 10 2 m 1 cm ))2( ( 7.0 cm )( 10 2 m 1 cm ))(i^)=(−0.286× 10 −4 N)(i^)

Conclusion:

Therefore, the force on segment 1, 2, 3 and 4 are −(2.5×10−5 N)(j^) , (1.00×10−4 N)(i^) , (2.5×10−5 N)(j^) and (−0.286×10−4 N)(i^) respectively.

(b)

To determine

The net force on the coil

(b)

Expert Solution

Answer to Problem 81P

The net force on the coil is (0.71×10−4 N)i^ .

Explanation of Solution

Formula used:

The expression for net force on the coil is given by,

Fnet=F1+F2+F3+F4

Calculation:

The net force on the coil is calculated as,

Fnet=F1+F2+F3+F4=−(2.5× 10 −5 N)(j^)+(( 1.00× 10 −4 N)( i ^))+(( 2.5× 10 −5 N)( j ^))+(−0.286× 10 −4 N)(i^)=(1.00× 10 −4 N)i^+(−0.286× 10 −4 N)(i^)=(0.71× 10 −4 N)i^

Conclusion:

Therefore, the net force on the coil is (0.71×10−4 N)i^ .

Answer 6

Chapter 27, Problem 81P

(a)

To determine

The force on each segment of rectangular coil.

(a)

Expert Solution

Answer to Problem 81P

The force on segment 1, 2, 3 and 4 are −(2.5×10−5 N)(j^) , (1.00×10−4 N)(i^) , (2.5×10−5 N)(j^) and (−0.286×10−4 N)(i^) respectively.

Explanation of Solution

Given:

The straight wire carries a current I1=20.0 A .

The coil carries a current I2=5.00 A .

The length of parallel sides are 5.00 cm and 10.0 cm .

Formula used:

The expression for force in first segment is given by,

F1=−μ0I1I22πlog(7.0 cm2.0 cm)(j^)

The expression for force in second segment is given by,

F2=−μ0I1I2l22π(2.0 cm)(i^)

The expression for force in third segment is given by,

F3=−F1

The expression for force in fourth segment is given by,

F4=−μ0I1I2l42π(7.0 cm)(i^)

Calculation:

The force in first segment is calculated as,

F1=−μ0I1I22πlog( 7.0 cm 2.0 cm)(j^)=−( 4π× 10 −7 N/ A 2 )( 20.0 A)( 5.00 A)2πlog( 7.0 cm 2.0 cm)(j^)=−(2.5× 10 −5 N)(j^)

The force in second segment is calculated as,

F2=−μ0I1I2l22π( 2.0 cm)(i^)=−( 4π× 10 −7 N/ A 2 )( 20.0 A)( 5.00 A)( ( 10 cm )( 10 2 m 1 cm ))2π( ( 2.0 cm )( 10 2 m 1 cm ))l(i^)=(1.00× 10 −4 N)(i^)

The expression for force in third segment is given by,

F3=−F1=−(−2.5× 10 −5 N)(j^)=(2.5× 10 −5 N)(j^)

The expression for force in fourth segment is given by,

F4=−μ0I1I2l42π( 7.0 cm)(i^)=−( 4π× 10 −7 N/ A 2 )( 20.0 A)( 5.00 A)( ( 10 cm )( 10 2 m 1 cm ))2( ( 7.0 cm )( 10 2 m 1 cm ))(i^)=(−0.286× 10 −4 N)(i^)

Conclusion:

Therefore, the force on segment 1, 2, 3 and 4 are −(2.5×10−5 N)(j^) , (1.00×10−4 N)(i^) , (2.5×10−5 N)(j^) and (−0.286×10−4 N)(i^) respectively.

Answer 7

Chapter 27, Problem 81P

(a)

To determine

The force on each segment of rectangular coil.

Answer 8

(a)

Expert Solution

Answer to Problem 81P

The force on segment 1, 2, 3 and 4 are −(2.5×10−5 N)(j^) , (1.00×10−4 N)(i^) , (2.5×10−5 N)(j^) and (−0.286×10−4 N)(i^) respectively.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

The straight wire carries a current I1=20.0 A .

The coil carries a current I2=5.00 A .

The length of parallel sides are 5.00 cm and 10.0 cm .

Formula used:

The expression for force in first segment is given by,

F1=−μ0I1I22πlog(7.0 cm2.0 cm)(j^)

The expression for force in second segment is given by,

F2=−μ0I1I2l22π(2.0 cm)(i^)

The expression for force in third segment is given by,

F3=−F1

The expression for force in fourth segment is given by,

F4=−μ0I1I2l42π(7.0 cm)(i^)

Calculation:

The force in first segment is calculated as,

F1=−μ0I1I22πlog( 7.0 cm 2.0 cm)(j^)=−( 4π× 10 −7 N/ A 2 )( 20.0 A)( 5.00 A)2πlog( 7.0 cm 2.0 cm)(j^)=−(2.5× 10 −5 N)(j^)

The force in second segment is calculated as,

F2=−μ0I1I2l22π( 2.0 cm)(i^)=−( 4π× 10 −7 N/ A 2 )( 20.0 A)( 5.00 A)( ( 10 cm )( 10 2 m 1 cm ))2π( ( 2.0 cm )( 10 2 m 1 cm ))l(i^)=(1.00× 10 −4 N)(i^)

The expression for force in third segment is given by,

F3=−F1=−(−2.5× 10 −5 N)(j^)=(2.5× 10 −5 N)(j^)

The expression for force in fourth segment is given by,

F4=−μ0I1I2l42π( 7.0 cm)(i^)=−( 4π× 10 −7 N/ A 2 )( 20.0 A)( 5.00 A)( ( 10 cm )( 10 2 m 1 cm ))2( ( 7.0 cm )( 10 2 m 1 cm ))(i^)=(−0.286× 10 −4 N)(i^)

Conclusion:

Therefore, the force on segment 1, 2, 3 and 4 are −(2.5×10−5 N)(j^) , (1.00×10−4 N)(i^) , (2.5×10−5 N)(j^) and (−0.286×10−4 N)(i^) respectively.

Answer 13

(b)

To determine

The net force on the coil

(b)

Expert Solution

Answer to Problem 81P

The net force on the coil is (0.71×10−4 N)i^ .

Explanation of Solution

Formula used:

The expression for net force on the coil is given by,

Fnet=F1+F2+F3+F4

Calculation:

The net force on the coil is calculated as,

Fnet=F1+F2+F3+F4=−(2.5× 10 −5 N)(j^)+(( 1.00× 10 −4 N)( i ^))+(( 2.5× 10 −5 N)( j ^))+(−0.286× 10 −4 N)(i^)=(1.00× 10 −4 N)i^+(−0.286× 10 −4 N)(i^)=(0.71× 10 −4 N)i^

Conclusion:

Therefore, the net force on the coil is (0.71×10−4 N)i^ .

Answer 14

(b)

To determine

The net force on the coil

Answer 15

(b)

Expert Solution

Answer to Problem 81P

The net force on the coil is (0.71×10−4 N)i^ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Formula used:

The expression for net force on the coil is given by,

Fnet=F1+F2+F3+F4

Calculation:

The net force on the coil is calculated as,

Fnet=F1+F2+F3+F4=−(2.5× 10 −5 N)(j^)+(( 1.00× 10 −4 N)( i ^))+(( 2.5× 10 −5 N)( j ^))+(−0.286× 10 −4 N)(i^)=(1.00× 10 −4 N)i^+(−0.286× 10 −4 N)(i^)=(0.71× 10 −4 N)i^