The magnetic force on a charge.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 26, Problem 15P

(a)

To determine

The magnetic force on a charge.

(a)

Expert Solution

Answer to Problem 15P

(3.804×10−6N)(−k^)

Explanation of Solution

Given:

Charge on the point particle =q=−3.64 nC =−3.64×10−9C

Velocity of the particle =v→=(2750ms)i^

Magnetic field =B→=(0.38 T)j^

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

F→=q(v→×B→)

Calculation:

Magnetic force on the particle is given as

F→=q(v→×B→)F→=(−3.64×10−9)((2750)i^×(0.38)j^)F→=(−3.64×10−9)(2750)(0.38)(i^×j^)F→=(−3.804×10−6)(k^)F→=(3.804×10−6N)(−k^)

Conclusion:

The magnetic force on the particle is (3.804×10−6N)(−k^) .

(b)

To determine

The magnetic force on a charge.

(b)

Expert Solution

Answer to Problem 15P

(7.51×10−6N)(−k^)

Explanation of Solution

Given:

Charge on the point particle =q=−3.64 nC =−3.64×10−9C

Velocity of the particle =v→=(2750ms)i^

Magnetic field =B→=(0.75 T)i^+(0.75 T)j^

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

F→=q(v→×B→)

Calculation:

Magnetic force on the particle is given as

F→=q(v→×B→)F→=(−3.64×10−9)((2750)i^×(0.75 i^+0.75 j^))F→=(−3.64×10−9)(2750)(0.75)(i^×i^)+(−3.64×10−9)(2750)(0.75)(i^×j^)F→=(−3.64×10−9)(2750)(0.75)(0)+(−3.64×10−9)(2750)(0.75)(k^)F→=0+(−7.51×10−6)(k^)F→=(7.51×10−6N)(−k^)

Conclusion:

The magnetic force on the particle is (7.51×10−6N)(−k^) .

(c)

To determine

The magnetic force on a charge.

(c)

Expert Solution

Answer to Problem 15P

0 N

Explanation of Solution

Given:

Charge on the point particle =q=−3.64 nC =−3.64×10−9C

Velocity of the particle =v→=(2750ms)i^

Magnetic field =B→=(0.65 T)i^

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

F→=q(v→×B→)

Calculation:

Magnetic force on the particle is given as

F→=q(v→×B→)F→=(−3.64×10−9)((2750)i^×(0.65)i^)F→=(−3.64×10−9)(2750)(0.65)(i^×i^)F→=(−3.64×10−9)(2750)(0.65)(0)F→=0 N

Conclusion:

The magnetic force on the particle is 0 N .

(d)

To determine

The magnetic force on a charge.

(d)

Expert Solution

Answer to Problem 15P

(7.51×10−6N)(j^)

Explanation of Solution

Given:

Charge on the point particle =q=−3.64 nC =−3.64×10−9C

Velocity of the particle =v→=(2750ms)i^

Magnetic field =B→=(0.75 T)i^+(0.75 T)k^

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

F→=q(v→×B→)

Calculation:

Magnetic force on the particle is given as

F→=q(v→×B→)F→=(−3.64×10−9)((2750)i^×(0.75 i^+0.75 k^))F→=(−3.64×10−9)(2750)(0.75)(i^×i^)+(−3.64×10−9)(2750)(0.75)(i^×k^)F→=(−3.64×10−9)(2750)(0.75)(0)+(−3.64×10−9)(2750)(0.75)(−j^)F→=0+(−7.51×10−6)(−j^)F→=(7.51×10−6N)(j^)

Conclusion:

The magnetic force on the particle is (7.51×10−6N)(j^) .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

A fluid with density 263 kg/m3 flows through a pipe of varying diameter and height. At location 1 the flow speed is 13.5 m/s and the diameter of the pipe is 7.4 cm down to location 2 the pipe diameter is 16.9 cm. Location 1 is 6.3 meters higher than location 2. What is the difference in pressure P2 - P1? Using units in Pascals and use g = 9.81 m/s2.

The kitchen had a temperature 46 degrees Fahrenheit and was converted it to Kelvin. What is the correct number for this temperature (46 F) on the Kelvin scale?

Water is traveling at a speed of 0.65 m/s through a pipe with a cross-section radius of 0.23 meters. The water enters a section of pipe that has a smaller radius, only 0.11 meters. What is the speed of the water traveling in this narrower section of pipe?

Answer 2

Question

Chapter 26, Problem 15P

(a)

To determine

The magnetic force on a charge.

(a)

Expert Solution

Answer to Problem 15P

(3.804×10−6N)(−k^)

Explanation of Solution

Given:

Charge on the point particle =q=−3.64 nC =−3.64×10−9C

Velocity of the particle =v→=(2750ms)i^

Magnetic field =B→=(0.38 T)j^

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

F→=q(v→×B→)

Calculation:

Magnetic force on the particle is given as

F→=q(v→×B→)F→=(−3.64×10−9)((2750)i^×(0.38)j^)F→=(−3.64×10−9)(2750)(0.38)(i^×j^)F→=(−3.804×10−6)(k^)F→=(3.804×10−6N)(−k^)

Conclusion:

The magnetic force on the particle is (3.804×10−6N)(−k^) .

(b)

To determine

The magnetic force on a charge.

(b)

Expert Solution

Answer to Problem 15P

(7.51×10−6N)(−k^)

Explanation of Solution

Given:

Charge on the point particle =q=−3.64 nC =−3.64×10−9C

Velocity of the particle =v→=(2750ms)i^

Magnetic field =B→=(0.75 T)i^+(0.75 T)j^

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

F→=q(v→×B→)

Calculation:

Magnetic force on the particle is given as

F→=q(v→×B→)F→=(−3.64×10−9)((2750)i^×(0.75 i^+0.75 j^))F→=(−3.64×10−9)(2750)(0.75)(i^×i^)+(−3.64×10−9)(2750)(0.75)(i^×j^)F→=(−3.64×10−9)(2750)(0.75)(0)+(−3.64×10−9)(2750)(0.75)(k^)F→=0+(−7.51×10−6)(k^)F→=(7.51×10−6N)(−k^)

Conclusion:

The magnetic force on the particle is (7.51×10−6N)(−k^) .

(c)

To determine

The magnetic force on a charge.

(c)

Expert Solution

Answer to Problem 15P

0 N

Explanation of Solution

Given:

Charge on the point particle =q=−3.64 nC =−3.64×10−9C

Velocity of the particle =v→=(2750ms)i^

Magnetic field =B→=(0.65 T)i^

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

F→=q(v→×B→)

Calculation:

Magnetic force on the particle is given as

F→=q(v→×B→)F→=(−3.64×10−9)((2750)i^×(0.65)i^)F→=(−3.64×10−9)(2750)(0.65)(i^×i^)F→=(−3.64×10−9)(2750)(0.65)(0)F→=0 N

Conclusion:

The magnetic force on the particle is 0 N .

(d)

To determine

The magnetic force on a charge.

(d)

Expert Solution

Answer to Problem 15P

(7.51×10−6N)(j^)

Explanation of Solution

Given:

Charge on the point particle =q=−3.64 nC =−3.64×10−9C

Velocity of the particle =v→=(2750ms)i^

Magnetic field =B→=(0.75 T)i^+(0.75 T)k^

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

F→=q(v→×B→)

Calculation:

Magnetic force on the particle is given as

F→=q(v→×B→)F→=(−3.64×10−9)((2750)i^×(0.75 i^+0.75 k^))F→=(−3.64×10−9)(2750)(0.75)(i^×i^)+(−3.64×10−9)(2750)(0.75)(i^×k^)F→=(−3.64×10−9)(2750)(0.75)(0)+(−3.64×10−9)(2750)(0.75)(−j^)F→=0+(−7.51×10−6)(−j^)F→=(7.51×10−6N)(j^)

Conclusion:

The magnetic force on the particle is (7.51×10−6N)(j^) .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 26, Problem 15P

(a)

To determine

The magnetic force on a charge.

(a)

Expert Solution

Answer to Problem 15P

(3.804×10−6N)(−k^)

Explanation of Solution

Given:

Charge on the point particle =q=−3.64 nC =−3.64×10−9C

Velocity of the particle =v→=(2750ms)i^

Magnetic field =B→=(0.38 T)j^

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

F→=q(v→×B→)

Calculation:

Magnetic force on the particle is given as

F→=q(v→×B→)F→=(−3.64×10−9)((2750)i^×(0.38)j^)F→=(−3.64×10−9)(2750)(0.38)(i^×j^)F→=(−3.804×10−6)(k^)F→=(3.804×10−6N)(−k^)

Conclusion:

The magnetic force on the particle is (3.804×10−6N)(−k^) .

(b)

To determine

The magnetic force on a charge.

(b)

Expert Solution

Answer to Problem 15P

(7.51×10−6N)(−k^)

Explanation of Solution

Given:

Charge on the point particle =q=−3.64 nC =−3.64×10−9C

Velocity of the particle =v→=(2750ms)i^

Magnetic field =B→=(0.75 T)i^+(0.75 T)j^

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

F→=q(v→×B→)

Calculation:

Magnetic force on the particle is given as

F→=q(v→×B→)F→=(−3.64×10−9)((2750)i^×(0.75 i^+0.75 j^))F→=(−3.64×10−9)(2750)(0.75)(i^×i^)+(−3.64×10−9)(2750)(0.75)(i^×j^)F→=(−3.64×10−9)(2750)(0.75)(0)+(−3.64×10−9)(2750)(0.75)(k^)F→=0+(−7.51×10−6)(k^)F→=(7.51×10−6N)(−k^)

Conclusion:

The magnetic force on the particle is (7.51×10−6N)(−k^) .

(c)

To determine

The magnetic force on a charge.

(c)

Expert Solution

Answer to Problem 15P

0 N

Explanation of Solution

Given:

Charge on the point particle =q=−3.64 nC =−3.64×10−9C

Velocity of the particle =v→=(2750ms)i^

Magnetic field =B→=(0.65 T)i^

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

F→=q(v→×B→)

Calculation:

Magnetic force on the particle is given as

F→=q(v→×B→)F→=(−3.64×10−9)((2750)i^×(0.65)i^)F→=(−3.64×10−9)(2750)(0.65)(i^×i^)F→=(−3.64×10−9)(2750)(0.65)(0)F→=0 N

Conclusion:

The magnetic force on the particle is 0 N .

(d)

To determine

The magnetic force on a charge.

(d)

Expert Solution

Answer to Problem 15P

(7.51×10−6N)(j^)

Explanation of Solution

Given:

Charge on the point particle =q=−3.64 nC =−3.64×10−9C

Velocity of the particle =v→=(2750ms)i^

Magnetic field =B→=(0.75 T)i^+(0.75 T)k^

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

F→=q(v→×B→)

Calculation:

Magnetic force on the particle is given as

F→=q(v→×B→)F→=(−3.64×10−9)((2750)i^×(0.75 i^+0.75 k^))F→=(−3.64×10−9)(2750)(0.75)(i^×i^)+(−3.64×10−9)(2750)(0.75)(i^×k^)F→=(−3.64×10−9)(2750)(0.75)(0)+(−3.64×10−9)(2750)(0.75)(−j^)F→=0+(−7.51×10−6)(−j^)F→=(7.51×10−6N)(j^)

Conclusion:

The magnetic force on the particle is (7.51×10−6N)(j^) .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 26, Problem 15P

(a)

To determine

The magnetic force on a charge.

(a)

Expert Solution

Answer to Problem 15P

(3.804×10−6N)(−k^)

Explanation of Solution

Given:

Charge on the point particle =q=−3.64 nC =−3.64×10−9C

Velocity of the particle =v→=(2750ms)i^

Magnetic field =B→=(0.38 T)j^

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

F→=q(v→×B→)

Calculation:

Magnetic force on the particle is given as

F→=q(v→×B→)F→=(−3.64×10−9)((2750)i^×(0.38)j^)F→=(−3.64×10−9)(2750)(0.38)(i^×j^)F→=(−3.804×10−6)(k^)F→=(3.804×10−6N)(−k^)

Conclusion:

The magnetic force on the particle is (3.804×10−6N)(−k^) .

(b)

To determine

The magnetic force on a charge.

(b)

Expert Solution

Answer to Problem 15P

(7.51×10−6N)(−k^)

Explanation of Solution

Given:

Charge on the point particle =q=−3.64 nC =−3.64×10−9C

Velocity of the particle =v→=(2750ms)i^

Magnetic field =B→=(0.75 T)i^+(0.75 T)j^

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

F→=q(v→×B→)

Calculation:

Magnetic force on the particle is given as

F→=q(v→×B→)F→=(−3.64×10−9)((2750)i^×(0.75 i^+0.75 j^))F→=(−3.64×10−9)(2750)(0.75)(i^×i^)+(−3.64×10−9)(2750)(0.75)(i^×j^)F→=(−3.64×10−9)(2750)(0.75)(0)+(−3.64×10−9)(2750)(0.75)(k^)F→=0+(−7.51×10−6)(k^)F→=(7.51×10−6N)(−k^)

Conclusion:

The magnetic force on the particle is (7.51×10−6N)(−k^) .

(c)

To determine

The magnetic force on a charge.

(c)

Expert Solution

Answer to Problem 15P

0 N

Explanation of Solution

Given:

Charge on the point particle =q=−3.64 nC =−3.64×10−9C

Velocity of the particle =v→=(2750ms)i^

Magnetic field =B→=(0.65 T)i^

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

F→=q(v→×B→)

Calculation:

Magnetic force on the particle is given as

F→=q(v→×B→)F→=(−3.64×10−9)((2750)i^×(0.65)i^)F→=(−3.64×10−9)(2750)(0.65)(i^×i^)F→=(−3.64×10−9)(2750)(0.65)(0)F→=0 N

Conclusion:

The magnetic force on the particle is 0 N .

(d)

To determine

The magnetic force on a charge.

(d)

Expert Solution

Answer to Problem 15P

(7.51×10−6N)(j^)

Explanation of Solution

Given:

Charge on the point particle =q=−3.64 nC =−3.64×10−9C

Velocity of the particle =v→=(2750ms)i^

Magnetic field =B→=(0.75 T)i^+(0.75 T)k^

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

F→=q(v→×B→)

Calculation:

Magnetic force on the particle is given as

F→=q(v→×B→)F→=(−3.64×10−9)((2750)i^×(0.75 i^+0.75 k^))F→=(−3.64×10−9)(2750)(0.75)(i^×i^)+(−3.64×10−9)(2750)(0.75)(i^×k^)F→=(−3.64×10−9)(2750)(0.75)(0)+(−3.64×10−9)(2750)(0.75)(−j^)F→=0+(−7.51×10−6)(−j^)F→=(7.51×10−6N)(j^)

Conclusion:

The magnetic force on the particle is (7.51×10−6N)(j^) .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 26, Problem 15P

(a)

To determine

The magnetic force on a charge.

(a)

Expert Solution

Answer to Problem 15P

(3.804×10−6N)(−k^)

Explanation of Solution

Given:

Charge on the point particle =q=−3.64 nC =−3.64×10−9C

Velocity of the particle =v→=(2750ms)i^

Magnetic field =B→=(0.38 T)j^

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

F→=q(v→×B→)

Calculation:

Magnetic force on the particle is given as

F→=q(v→×B→)F→=(−3.64×10−9)((2750)i^×(0.38)j^)F→=(−3.64×10−9)(2750)(0.38)(i^×j^)F→=(−3.804×10−6)(k^)F→=(3.804×10−6N)(−k^)

Conclusion:

The magnetic force on the particle is (3.804×10−6N)(−k^) .

(b)

To determine

The magnetic force on a charge.

(b)

Expert Solution

Answer to Problem 15P

(7.51×10−6N)(−k^)

Explanation of Solution

Given:

Charge on the point particle =q=−3.64 nC =−3.64×10−9C

Velocity of the particle =v→=(2750ms)i^

Magnetic field =B→=(0.75 T)i^+(0.75 T)j^

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

F→=q(v→×B→)

Calculation:

Magnetic force on the particle is given as

F→=q(v→×B→)F→=(−3.64×10−9)((2750)i^×(0.75 i^+0.75 j^))F→=(−3.64×10−9)(2750)(0.75)(i^×i^)+(−3.64×10−9)(2750)(0.75)(i^×j^)F→=(−3.64×10−9)(2750)(0.75)(0)+(−3.64×10−9)(2750)(0.75)(k^)F→=0+(−7.51×10−6)(k^)F→=(7.51×10−6N)(−k^)

Conclusion:

The magnetic force on the particle is (7.51×10−6N)(−k^) .

(c)

To determine

The magnetic force on a charge.

(c)

Expert Solution

Answer to Problem 15P

0 N

Explanation of Solution

Given:

Charge on the point particle =q=−3.64 nC =−3.64×10−9C

Velocity of the particle =v→=(2750ms)i^

Magnetic field =B→=(0.65 T)i^

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

F→=q(v→×B→)

Calculation:

Magnetic force on the particle is given as

F→=q(v→×B→)F→=(−3.64×10−9)((2750)i^×(0.65)i^)F→=(−3.64×10−9)(2750)(0.65)(i^×i^)F→=(−3.64×10−9)(2750)(0.65)(0)F→=0 N

Conclusion:

The magnetic force on the particle is 0 N .

(d)

To determine

The magnetic force on a charge.

(d)

Expert Solution

Answer to Problem 15P

(7.51×10−6N)(j^)

Explanation of Solution

Given:

Charge on the point particle =q=−3.64 nC =−3.64×10−9C

Velocity of the particle =v→=(2750ms)i^

Magnetic field =B→=(0.75 T)i^+(0.75 T)k^

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

F→=q(v→×B→)

Calculation:

Magnetic force on the particle is given as

F→=q(v→×B→)F→=(−3.64×10−9)((2750)i^×(0.75 i^+0.75 k^))F→=(−3.64×10−9)(2750)(0.75)(i^×i^)+(−3.64×10−9)(2750)(0.75)(i^×k^)F→=(−3.64×10−9)(2750)(0.75)(0)+(−3.64×10−9)(2750)(0.75)(−j^)F→=0+(−7.51×10−6)(−j^)F→=(7.51×10−6N)(j^)

Conclusion:

The magnetic force on the particle is (7.51×10−6N)(j^) .

Answer 6

Chapter 26, Problem 15P

(a)

To determine

The magnetic force on a charge.

(a)

Expert Solution

Answer to Problem 15P

(3.804×10−6N)(−k^)

Explanation of Solution

Given:

Charge on the point particle =q=−3.64 nC =−3.64×10−9C

Velocity of the particle =v→=(2750ms)i^

Magnetic field =B→=(0.38 T)j^

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

F→=q(v→×B→)

Calculation:

Magnetic force on the particle is given as

F→=q(v→×B→)F→=(−3.64×10−9)((2750)i^×(0.38)j^)F→=(−3.64×10−9)(2750)(0.38)(i^×j^)F→=(−3.804×10−6)(k^)F→=(3.804×10−6N)(−k^)

Conclusion:

The magnetic force on the particle is (3.804×10−6N)(−k^) .

Answer 7

Chapter 26, Problem 15P

(a)

To determine

The magnetic force on a charge.

Answer 8

(a)

Expert Solution

Answer to Problem 15P

(3.804×10−6N)(−k^)

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

Charge on the point particle =q=−3.64 nC =−3.64×10−9C

Velocity of the particle =v→=(2750ms)i^

Magnetic field =B→=(0.38 T)j^

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

F→=q(v→×B→)

Calculation:

Magnetic force on the particle is given as

F→=q(v→×B→)F→=(−3.64×10−9)((2750)i^×(0.38)j^)F→=(−3.64×10−9)(2750)(0.38)(i^×j^)F→=(−3.804×10−6)(k^)F→=(3.804×10−6N)(−k^)

Conclusion:

The magnetic force on the particle is (3.804×10−6N)(−k^) .

Answer 13

(b)

To determine

The magnetic force on a charge.

(b)

Expert Solution

Answer to Problem 15P

(7.51×10−6N)(−k^)

Explanation of Solution

Given:

Charge on the point particle =q=−3.64 nC =−3.64×10−9C

Velocity of the particle =v→=(2750ms)i^

Magnetic field =B→=(0.75 T)i^+(0.75 T)j^

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

F→=q(v→×B→)

Calculation:

Magnetic force on the particle is given as

F→=q(v→×B→)F→=(−3.64×10−9)((2750)i^×(0.75 i^+0.75 j^))F→=(−3.64×10−9)(2750)(0.75)(i^×i^)+(−3.64×10−9)(2750)(0.75)(i^×j^)F→=(−3.64×10−9)(2750)(0.75)(0)+(−3.64×10−9)(2750)(0.75)(k^)F→=0+(−7.51×10−6)(k^)F→=(7.51×10−6N)(−k^)

Conclusion:

The magnetic force on the particle is (7.51×10−6N)(−k^) .

Answer 14

(b)

To determine

The magnetic force on a charge.

Answer 15

(b)

Expert Solution

Answer to Problem 15P

(7.51×10−6N)(−k^)

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given:

Charge on the point particle =q=−3.64 nC =−3.64×10−9C

Velocity of the particle =v→=(2750ms)i^

Magnetic field =B→=(0.75 T)i^+(0.75 T)j^

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

F→=q(v→×B→)

Calculation:

Magnetic force on the particle is given as

F→=q(v→×B→)F→=(−3.64×10−9)((2750)i^×(0.75 i^+0.75 j^))F→=(−3.64×10−9)(2750)(0.75)(i^×i^)+(−3.64×10−9)(2750)(0.75)(i^×j^)F→=(−3.64×10−9)(2750)(0.75)(0)+(−3.64×10−9)(2750)(0.75)(k^)F→=0+(−7.51×10−6)(k^)F→=(7.51×10−6N)(−k^)

Conclusion:

The magnetic force on the particle is (7.51×10−6N)(−k^) .

Answer 20

(c)

To determine

The magnetic force on a charge.

(c)

Expert Solution

Answer to Problem 15P

0 N

Explanation of Solution

Given:

Charge on the point particle =q=−3.64 nC =−3.64×10−9C

Velocity of the particle =v→=(2750ms)i^

Magnetic field =B→=(0.65 T)i^

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

F→=q(v→×B→)

Calculation:

Magnetic force on the particle is given as

F→=q(v→×B→)F→=(−3.64×10−9)((2750)i^×(0.65)i^)F→=(−3.64×10−9)(2750)(0.65)(i^×i^)F→=(−3.64×10−9)(2750)(0.65)(0)F→=0 N

Conclusion:

The magnetic force on the particle is 0 N .

Answer 21

(c)

To determine

The magnetic force on a charge.

Answer 22

(c)

Expert Solution

Answer to Problem 15P

0 N

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given:

Charge on the point particle =q=−3.64 nC =−3.64×10−9C

Velocity of the particle =v→=(2750ms)i^

Magnetic field =B→=(0.65 T)i^

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

F→=q(v→×B→)

Calculation:

Magnetic force on the particle is given as

F→=q(v→×B→)F→=(−3.64×10−9)((2750)i^×(0.65)i^)F→=(−3.64×10−9)(2750)(0.65)(i^×i^)F→=(−3.64×10−9)(2750)(0.65)(0)F→=0 N

Conclusion:

The magnetic force on the particle is 0 N .

Answer 27

(d)

To determine

The magnetic force on a charge.

(d)

Expert Solution

Answer to Problem 15P

(7.51×10−6N)(j^)

Explanation of Solution

Given:

Charge on the point particle =q=−3.64 nC =−3.64×10−9C

Velocity of the particle =v→=(2750ms)i^

Magnetic field =B→=(0.75 T)i^+(0.75 T)k^

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

F→=q(v→×B→)

Calculation:

Magnetic force on the particle is given as

F→=q(v→×B→)F→=(−3.64×10−9)((2750)i^×(0.75 i^+0.75 k^))F→=(−3.64×10−9)(2750)(0.75)(i^×i^)+(−3.64×10−9)(2750)(0.75)(i^×k^)F→=(−3.64×10−9)(2750)(0.75)(0)+(−3.64×10−9)(2750)(0.75)(−j^)F→=0+(−7.51×10−6)(−j^)F→=(7.51×10−6N)(j^)

Conclusion:

The magnetic force on the particle is (7.51×10−6N)(j^) .

Answer 28

(d)

To determine

The magnetic force on a charge.

Answer 29

(d)

Expert Solution

Answer to Problem 15P

(7.51×10−6N)(j^)

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Given:

Charge on the point particle =q=−3.64 nC =−3.64×10−9C

Velocity of the particle =v→=(2750ms)i^

Magnetic field =B→=(0.75 T)i^+(0.75 T)k^

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

F→=q(v→×B→)

Calculation:

Magnetic force on the particle is given as

F→=q(v→×B→)F→=(−3.64×10−9)((2750)i^×(0.75 i^+0.75 k^))F→=(−3.64×10−9)(2750)(0.75)(i^×i^)+(−3.64×10−9)(2750)(0.75)(i^×k^)F→=(−3.64×10−9)(2750)(0.75)(0)+(−3.64×10−9)(2750)(0.75)(−j^)F→=0+(−7.51×10−6)(−j^)F→=(7.51×10−6N)(j^)