Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Question
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Chapter 27, Problem 81P

(a)

To determine

The force on each segment of rectangular coil.

(a)

Expert Solution
Check Mark

Answer to Problem 81P

The force on segment 1, 2, 3 and 4 are (2.5×105N)(j^) , (1.00×104N)(i^) , (2.5×105N)(j^) and (0.286×104N)(i^) respectively.

Explanation of Solution

Given:

The straight wire carries a current I1=20.0A .

The coil carries a current I2=5.00A .

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

Formula used:

The expression for force in first segment is given by,

  F1=μ0I1I22πlog(7.0cm2.0cm)(j^)

The expression for force in second segment is given by,

  F2=μ0I1I2l22π(2.0cm)(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(7.0cm)(i^)

Calculation:

The force in first segment is calculated as,

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

The force in second segment is calculated as,

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

The expression for force in third segment is given by,

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

The expression for force in fourth segment is given by,

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

Conclusion:

Therefore, the force on segment 1, 2, 3 and 4 are (2.5×105N)(j^) , (1.00×104N)(i^) , (2.5×105N)(j^) and (0.286×104N)(i^) respectively.

(b)

To determine

The net force on the coil

(b)

Expert Solution
Check Mark

Answer to Problem 81P

The net force on the coil is (0.71×104N)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 5N)(j^)+(( 1.00× 10 4 N)( i ^))+(( 2.5× 10 5 N)( j ^))+(0.286× 10 4N)(i^)=(1.00× 10 4N)i^+(0.286× 10 4N)(i^)=(0.71× 10 4N)i^

Conclusion:

Therefore, the net force on the coil is (0.71×104N)i^ .

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Chapter 27 Solutions

Physics for Scientists and Engineers

Ch. 27 - Prob. 11PCh. 27 - Prob. 12PCh. 27 - Prob. 13PCh. 27 - Prob. 14PCh. 27 - Prob. 15PCh. 27 - Prob. 16PCh. 27 - Prob. 17PCh. 27 - Prob. 18PCh. 27 - Prob. 19PCh. 27 - Prob. 20PCh. 27 - Prob. 21PCh. 27 - Prob. 22PCh. 27 - Prob. 23PCh. 27 - Prob. 24PCh. 27 - Prob. 25PCh. 27 - Prob. 26PCh. 27 - Prob. 27PCh. 27 - Prob. 28PCh. 27 - Prob. 29PCh. 27 - Prob. 30PCh. 27 - Prob. 31PCh. 27 - Prob. 32PCh. 27 - Prob. 33PCh. 27 - Prob. 34PCh. 27 - Prob. 35PCh. 27 - Prob. 36PCh. 27 - Prob. 37PCh. 27 - Prob. 38PCh. 27 - Prob. 39PCh. 27 - Prob. 40PCh. 27 - Prob. 41PCh. 27 - Prob. 42PCh. 27 - Prob. 43PCh. 27 - Prob. 44PCh. 27 - Prob. 45PCh. 27 - Prob. 46PCh. 27 - Prob. 47PCh. 27 - Prob. 48PCh. 27 - Prob. 49PCh. 27 - Prob. 50PCh. 27 - Prob. 51PCh. 27 - Prob. 52PCh. 27 - Prob. 53PCh. 27 - Prob. 54PCh. 27 - Prob. 55PCh. 27 - Prob. 56PCh. 27 - Prob. 57PCh. 27 - Prob. 58PCh. 27 - Prob. 59PCh. 27 - Prob. 60PCh. 27 - Prob. 61PCh. 27 - Prob. 62PCh. 27 - Prob. 63PCh. 27 - Prob. 64PCh. 27 - Prob. 65PCh. 27 - Prob. 66PCh. 27 - Prob. 67PCh. 27 - Prob. 68PCh. 27 - Prob. 69PCh. 27 - Prob. 70PCh. 27 - Prob. 71PCh. 27 - Prob. 72PCh. 27 - Prob. 73PCh. 27 - Prob. 74PCh. 27 - Prob. 75PCh. 27 - Prob. 76PCh. 27 - Prob. 77PCh. 27 - Prob. 78PCh. 27 - Prob. 79PCh. 27 - Prob. 80PCh. 27 - Prob. 81PCh. 27 - Prob. 82PCh. 27 - Prob. 83PCh. 27 - Prob. 84PCh. 27 - Prob. 85PCh. 27 - Prob. 86PCh. 27 - Prob. 87PCh. 27 - Prob. 88PCh. 27 - Prob. 89PCh. 27 - Prob. 90PCh. 27 - Prob. 91PCh. 27 - Prob. 92PCh. 27 - Prob. 93PCh. 27 - Prob. 94PCh. 27 - Prob. 95PCh. 27 - Prob. 96PCh. 27 - Prob. 97PCh. 27 - Prob. 98P
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