FlipIt for College Physics (Algebra Version - Six Months Access)
FlipIt for College Physics (Algebra Version - Six Months Access)
17th Edition
ISBN: 9781319032432
Author: Todd Ruskell
Publisher: W.H. Freeman & Co
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Chapter 20, Problem 25QAP
To determine

(a)

Magnitude and direction of the force on each side of the coil in problem (24) for situation a), b) and c)

Expert Solution
Check Mark

Answer to Problem 25QAP

Each side of b) situation have current of 79.2 mA

In top side, force towards bottom

In bottom side, force towards top

In left side, force towards right

In right side, force towards left

In situation a) and b) there is no induced current, so there is no force

Explanation of Solution

Given info:

  induced current insituation a) = 0 Asituation b) = 0.044 Asituation c) = 0 AArea of the coil= 0.01 m2Number of turns = 30magnetic field = 0.6 T

Formula used:

  F=NIlBF= magnetic forceB= magnetic fieldl= side lengthI=  currentN=number of turns

Calculation:

There is no induced current in a) and c) so there is no magnetic force in a) and c)

In b), there is an induced current, so the force on sides

  According to the Lenz's law, current flows counterclockwise to produce a field that goes in to the page

  Top side,F=NIlBF=30*0.044 A*0.6 T*0.1 mF=79.2 mNDirection =towards south

  Bottom side,F=NIlBF=30*0.044 A*0.6 T*0.1 mF=79.2 mNDirection =towards north

  left side,F=NIlBF=30*0.044 A*0.6 T*0.1 mF=79.2 mNDirection =towards rightright side,F=NIlBF=30*0.044 A*0.6 T*0.1 mF=79.2 mNDirection =towards left

Conclusion:

Each side of b) situation have current of 79.2 mA

In top side, force towards bottom

In bottom side, force towards top

In left side, force towards right

In right side, force towards left

In situation a) and b) there is no induced current, so there is no force

To determine

(b)

Magnitude and direction of the force on each side of the coil when it enters from the left

Expert Solution
Check Mark

Answer to Problem 25QAP

Each side of have current of 79.2 mA

In top side, force towards bottom

In bottom side, force towards top

In left side, force towards right

In right side, force towards left

Explanation of Solution

Given info:

  induced current insituation a) = 0 Asituation b) = 0.044 Asituation c) = 0 AArea of the coil= 0.01 m2Number of turns = 30magnetic field = 0.6 T

Formula used:

  F=NIlBF= magnetic forceB= magnetic fieldl= side lengthI=  currentN=number of turns

Calculation:

As the loop enters the field there will be an induced current in the loop

  According to the Lenz's law, current flows counterclockwise to produce a field that comes out of the page

  Top side,F=NIlBF=30*0.044 A*0.6 T*0.1 mF=79.2 mNDirection =towards south

  Bottom side,F=NIlBF=30*0.044 A*0.6 T*0.1 mF=79.2 mNDirection =towards north

  left side,F=NIlBF=30*0.044 A*0.6 T*0.1 mF=79.2 mNDirection =towards rightright side,F=NIlBF=30*0.044 A*0.6 T*0.1 mF=79.2 mNDirection =towards left

Conclusion:

Each side have current of 79.2 mA

In top side, force towards bottom

In bottom side, force towards top

In left side, force towards right

In right side, force towards left

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