EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
9th Edition
ISBN: 9781305804463
Author: Jewett
Publisher: CENGAGE LEARNING - CONSIGNMENT
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Chapter 29, Problem 29.44P

In Figure P28.28, the cube is 40.0 cm on each edge. Four straight segments of wire—ab, bc, cd, and da—form a closed loop that carries a current I = 5.00 A in the direction shown. A uniform magnetic field of magnitude B = 0.020 0 T is in the positive y direction. Determine the magnetic force vector on (a) ab, (b) bc, (c) cd, and (d) da. (c) Explain how you could find the force exerted on the fourth of these segments from the forces on the other three, without further calculation involving the magnetic field.

Figure P28.28

Chapter 29, Problem 29.44P, In Figure P28.28, the cube is 40.0 cm on each edge. Four straight segments of wireab, bc, cd, and

(a)

Expert Solution
Check Mark
To determine
The magnetic force vector on ab .

Answer to Problem 29.44P

The magnetic force vector on ab is 0.

Explanation of Solution

Given info: The length of each edge of the cube is 40.0cm , the electric current is 5.00A and the magnitude of magnetic field is 0.020T .

The formula for the magnetic force is,

F=BIlsinθ

Here,

B is the magnitude of magnetic field.

I is the intensity of electric current.

l is the length of the cube side ab .

θ is the angle between length and magnetic field.

As the current flows down and the magnetic from point b is upwards so the angle θ is 180° .

Substitute 0.020T for B , 5.00A for I , 180° for θ and 40.0cm for l in the above equation to find the value of F .

F=(0.020T)(5.00A)(40.0cm×102m1cm)sin180°=0

The magnitude of the force is zero because an equal and opposite force cancels it due to which the magnetic force of ab have no direction.

Conclusion:

Therefore, magnetic force vector on ab is 0.

(b)

Expert Solution
Check Mark
To determine
The magnetic force vector on bc .

Answer to Problem 29.44P

The magnetic force vector on bc is 0.04i^N .

Explanation of Solution

Given info: The length of each edge of the cube is 40.0cm , the electric current is 5.00A and the magnitude of magnetic field is 0.020T

The formula for the magnetic force is,

F=BIlsinθ

Here,

B is the magnitude of magnetic field.

I is the intensity of electric current.

l is the length of the cube side bc .

θ is the angle between length and magnetic field.

As the current flows down and the magnetic from point c is perpendicular so the angle θ is 90° .

Substitute 0.020T for B , 5.00A for I , 90° for θ and 40.0cm for l in the above equation to find the value of F .

F=(0.020T)(5.00A)(40.0cm×102m1cm)sin90°=0.04N

By using the Flemings right hand rule, the thumb points towards the x axis direction. Thus, the direction of the magnetic force is in x -direction.

Conclusion:

Therefore, the magnetic force vector on bc is 0.04i^N .

(c)

Expert Solution
Check Mark
To determine
The magnetic force vector on cd .

Answer to Problem 29.44P

The magnetic force vector on cd is 0.04k^N .

Explanation of Solution

Given info: The length of each edge of the cube is 40.0cm , the electric current is 5.00A and the magnitude of magnetic field is 0.020T

The formula for the magnetic force is,

F=BIlsinθ

Here,

B is the magnitude of magnetic field.

I is the intensity of electric current.

l is the length of the cube side cd .

θ is the angle between length and magnetic field.

Use Pythagoras theorem to find l ,

l=a2+a2=2a2

Substitute 40.0cm for a in the above equation to find the value of l .

l=2×40.0cm2=57cm

As the current flows at an angle and the magnetic from point b is perpendicular so the angle θ is 45° .

Substitute 0.020T for B , 5.00A for I , 45° for θ and 57.0cm for l in the above equation to find the value of F .

F=(0.020T)(5.00A)(57.0cm×102m1cm)sin45°=0.04N

By using the Flemings right hand rule, the thumb points towards the z axis direction. Thus, the direction of the magnetic force is in z direction.

Conclusion:

Therefore, the magnetic force vector on cd is 0.04k^N .

(d)

Expert Solution
Check Mark
To determine
The magnetic force vector on da .

Answer to Problem 29.44P

The magnetic force vector on da is 0.057(12(i^+k^))N .

Explanation of Solution

Given info: The length of each edge of the cube is 40.0cm , the electric current is 5.00A and the magnitude of magnetic field is 0.020T

The formula for the magnetic force is,

F=BIlsinθ

Here,

B is the magnitude of magnetic field.

I is the intensity of electric current.

l is the length of the cube side da .

θ is the angle between length and magnetic field.

Use Pythagoras theorem to find l ,

l=a2+a2=2a2

Substitute 40.0cm for a in the above equation to find the value of l .

l=2×40.0cm2=57cm

As the current flows vertically and the magnetic from point b is perpendicular so the angle θ is 90° .

Substitute 0.020T for B , 5.00A for I , 90° for θ and 57.0cm for l in the above equation to find the value of F .

F=(0.020T)(5.00A)(57.0cm×102m1cm)sin90°=0.057N

By using the Flemings right hand rule, the thumb points towards the direction d

The direction of the force is,

d=cos45°i^+cos45°k^=(12(i^+k^))

Thus, the direction of the magnetic force is (12(i^+k^)) .

Conclusion:

Therefore, the magnetic force vector on da is 0.057(12(i^+k^))N .

(e)

Expert Solution
Check Mark
To determine
How to find the force exerted on the forth segment from the forces on the other three.

Answer to Problem 29.44P

The force exerted on the forth segment from the forces on the other three can be calculated by the parallelogram law of vectors.

Explanation of Solution

Given info: The length of each edge of the cube is 40.0cm , the electric current is 5.00A and the magnitude of magnetic field is 0.020T

By the parallelogram law of forces, when the forces on three of the arms of a parallelogram are provided then the magnitude of force on the forth arm is equal to the resultant of the other three forces.

According to the parallelogram law of vectors,

F4=F3+F2+F1

Here,

F4 is the force of the resultant.

F3 is the force on the third arm of the parallelogram.

F1 is the force on the first arm of the parallelogram.

F2 is the force on the second arm of the parallelogram.

Conclusion:

Therefore, the force exerted on the forth segment from the forces on the other three can be calculated by the parallelogram law of vectors.

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Chapter 29 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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