Physics
Physics
3rd Edition
ISBN: 9780073512150
Author: Alan Giambattista, Betty Richardson, Robert C. Richardson Dr.
Publisher: McGraw-Hill Education
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Chapter 19, Problem 50P

(a)

To determine

What is the magnetic force on each side of the loop if the magnetic field is 2.5T out of the page?

(a)

Expert Solution
Check Mark

Answer to Problem 50P

Magnetic force on the top side of loop is 0.75N in negative y direction, force on bottom side of loop is 0.75N in the positive y direction, force on left side of loop is 0.50N in the positive x direction, and force on the right side of the loop is 0.50N in the negative x direction.

Explanation of Solution

The schematic diagram showing the rectangular loop carrying current is given in figure 1 below. The direction of magnetic field is given to be out of the page. The four sides of the loop is also marked in the figure. Consider each side separately to find the magnitude and direction of force experienced in the loop.

Physics, Chapter 19, Problem 50P

Write the equation to find the force on the top side of the loop.

Ftop=ILtop×B (I)

Here, Ftop is the force on top side of loop, I is the current in the loop, Ltop is the length of the top side, B is the magnetic field vector

Write the equation to find the force on the bottom side of the loop.

Fbottom=ILbottom×B (II)

Here, Ftop is the force on bottom side of loop, I is the current in the loop, Ltop is the length of the bottom side, B is the magnetic field vector

Write the equation to find the force on the left side of the loop.

Fleft=ILleft×B (III)

Here, Fleft is the force on left side of loop, I is the current in the loop, Lleft is the length of the left side, B is the magnetic field vector

Write the equation to find the force on the right side of the loop.

Fright=ILright×B (IV)

Here, Fright is the force on right side of loop, I is the current in the loop, Lright is the length of the right side, B is the magnetic field vector

Conclusion

Substitute 1.0A for I , 0.300m right for Ltop , 2.5Tout of page for B in equation (I) to get Ftop

Ftop=(1.0A)(0.300mright)×(2.5Tout of page)=0.75N in the -y direction

Substitute 1.0A for I , 0.300m left for Ltop , 2.5Tout of page for B in equation (II) to get Fbottom

Fbottom=(1.0A)(0.300mleft)×(2.5Tout of page)=0.75N in the +y direction

Substitute 1.0A for I , 0.300m up for Ltop , 2.5Tout of page for B in equation (III) to get Fleft

Fleft=(1.0A)(0.300mup)×(2.5Tout of page)=0.50N in the +x direction

Substitute 1.0A for I , 0.300m down for Lright , 2.5Tout of page for B in equation (IV) to get Fright

Fright=(1.0A)(0.300mdown)×(2.5Tout of page)=0.50N in the -x direction

Therefore, Magnetic force on the top side of loop is 0.75N in negative y direction, force on bottom side of loop is 0.75N in the positive y direction, force on left side of loop is 0.50N in the positive x direction, and force on the right side of the loop is 0.50N in the negative x direction.

(b)

To determine

What is the net magnetic force on the loop?

(b)

Expert Solution
Check Mark

Answer to Problem 50P

The net magnetic force on the loop is 0N.

Explanation of Solution

The net force is the resultant sum of forces in the x and y directions.

Write the equation to find the net force in x direction.

Fnet,x=Fleft+Fright (I)

Here, Fnet,x is the net force in the x direction, Fleft is the force in the left side of loop, Fright is the force in the right side of the loop

Write the equation to find the net force in y direction.

Fnet,y=Ftop+Fbottom (II)

Here, Fnet,y is the net force in the y direction, Ftop is the force in the top side of loop, Fbottom is the force in the bottom side of the loop

Write the equation to find the sum of x and y component of forces.

Fnet=Fnet,x+Fnet,y (III)

Here, Fnet is the net force in the loop, Fnet,x is the net force in the x direction, Fnet,y is the net force in the y direction

Conclusion:

Substitute 0.50N for Fleft , and 0.50N for Fright in equation (I) to get Fnet,x

Fnet,x=0.50N-0.50N=0N

Substitute 0.75N for Fbottom , and 0.75N for Ftop in equation (II) to get Fnet,y

Fnet,y=0.75N-0.75N=0N

Substitute 0N for Fnet,x and Fnet,y in equation (III) to get Fnet

Fnet=0N+0N=0N

Therefore, The net magnetic force on the loop is 0N.

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Chapter 19 Solutions

Physics

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In the figure, four...Ch. 19 - Prob. 4MCQCh. 19 - Prob. 5MCQCh. 19 - Prob. 6MCQCh. 19 - Prob. 7MCQCh. 19 - Prob. 8MCQCh. 19 - Multiple-Choice Questions 6-9. A wire carries...Ch. 19 - Prob. 10MCQCh. 19 - 11. The magnetic forces that two parallel wires...Ch. 19 - Prob. 12MCQCh. 19 - 1. At which point in the diagram is the magnetic...Ch. 19 - 2. Draw vector arrows to indicate the direction...Ch. 19 - Problems 3-6. Sketch some magnetic field lines for...Ch. 19 - Prob. 4PCh. 19 - Prob. 5PCh. 19 - Problems 3–6. Sketch some magnetic field lines for...Ch. 19 - 7. Find the magnetic force exerted on an electron...Ch. 19 - 8. Find the magnetic force exerted on a proton...Ch. 19 - 9. A uniform magnetic field points north; its...Ch. 19 - 10. A uniform magnetic field points vertically...Ch. 19 - Problems 11-14. Several electrons move at speed...Ch. 19 - 12. Find the magnetic force on the electron at...Ch. 19 - 12. Find the magnetic force on the electron at...Ch. 19 - Problems 11-14. 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