COLLEGE PHYSICS
COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 19, Problem 78QAP
To determine

(a)

The magnitude of the magnetic field the cord produces midway between the two wires

Expert Solution
Check Mark

Answer to Problem 78QAP

Magnitude of the magnetic field the cord produces midway between the two wires= 1.36×104 T

Explanation of Solution

Given info:

  Voltage = 110 VThe electric power of the lamp = 75 Wtwo insulated parallel wires 4.0 mm apartEarth's magnetic field = 5×10-5 T

Formula used:

  P=VIV=potentialP=PowerI=current

  B=μ0I2πdB=Magnetic fieldμ0=permeability of free spaceI=Current in wired=Radius from the wire

Calculation:

  P=VI75=110×II=0.68 A

  B=μ0I2πdB=4π×107×0.682π×2×103B=0.68×104 TBoth B due to two wires are equal in magnitude and to same directionBTotal=2×0.68×104 TBTotal=1.36×104 T

Conclusion:

Magnitude of the magnetic field the cord produces midway between the two wires= 1.36×104 T

To determine

(b)

The magnitude of the magnetic field the cord produces 2.0 mm from one of the wires in the same plane in which the two wires lie

Expert Solution
Check Mark

Answer to Problem 78QAP

Magnitude of the magnetic field the cord produces 2.0 mm from one of the wires in the same plane in which the two wires lie= 0.45*104 T

Explanation of Solution

Given info:

  Voltage = 110 VThe electric power of the lamp = 75 Wtwo insulated parallel wires 4.0 mm apartEarth's magnetic field = 5*10-5 T

Formula used:

  P=VIV=potentialP=PowerI=current

  B=μ0I2πdB=Magnetic fieldμ0=permeability of free spaceI=Current in wired=Radius from the wire

Calculation:

For the wire on the right side d =2 mm

  B=μ0I2πdB=4π×107×0.682π×2×103B=0.68×104 T

Field is directed outwards.

For the wire on the left side d =2+4 mm=6 mm

  B=μ0I2πdB=4π*107*0.682π*6*103B=0.23*104 T

Field is directed inwards.

  Btotal=0.68*104 T0.23*104 T=0.45*104 T

Field is directed outwards.

Conclusion:

Magnitude of the magnetic field the cord produces 2.0 mm from one of the wires in the same plane in which the two wires lie= 0.45*104 T

To determine

(c)

Comparison of previous magnetic fields with earth's magnetic field

Expert Solution
Check Mark

Answer to Problem 78QAP

For case a),

  BBearth=2.72

For case b),

  BBearth=0.9

Explanation of Solution

Given info:

  Voltage = 110 VThe electric power of the lamp = 75 Wtwo insulated parallel wires 4.0 mm apartEarth's magnetic field = 5*10-5 T

Formula used:

  P=VIV=potentialP=PowerI=current

  B=μ0I2πdB=Magnetic fieldμ0=permeability of free spaceI=Current in wired=Radius from the wire

Calculation:

For case a),

  BBearth=1.36*104 T5*105 T=2.72

For case b),

  BBearth=0.45*104 T5*105 T=0.9

Conclusion:

For case a),

  BBearth=2.72

For case b),

  BBearth=0.9

To determine

(d)

Magnetic force do the two wires exert on one another

Expert Solution
Check Mark

Answer to Problem 78QAP

Force= 0.46*104 N

Force is very small

Explanation of Solution

Given info:

  Voltage = 110 VThe electric power of the lamp = 75 Wtwo insulated parallel wires 4.0 mm apartEarth's magnetic field = 5*10-5 T

Formula used:

  F=μ0I2l2πdB=Magnetic fieldμ0=permeability of free spaceI=Current in wired=Radius from the wirel=length of wire

Calculation:

Force on each wire.

  F=μ0I2l2πdF=4π*107*0.682*22π*4*103F=0.46*104 N

Force is very small

Conclusion:

Force= 0.46*104 N

Force is very small

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Chapter 19 Solutions

COLLEGE PHYSICS

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