The force on each segment of rectangular coil.

Question

Want to see more full solutions like this?

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^ .

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!

Students have asked these similar questions

Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.

A long wire and coil are located in one plane, and a current of 4A flows through the wire and a current of 10A passes through the coil. If the coil is in the form of a rectangle, its dimensions are 10 cm and 20 cm and the number of turns is 40, and the nearest long side is 10 cm away from the wire, then it is the force applied to one of the two short sides before The long wire is

None

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^ .

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 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^ .

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 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^ .

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 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^