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 small object has a charge of 2 C has a mass of 0.5 kg and a velocity of u= 3ax+4az m/s at a particular instant in time. If the point charge is moving through an electromagnetic field with E = 2ax+2ay V/m and B a = 4ax Wb/m^2, what is the total the acceleration of the object at the instant of time?

A proton moving at speed v = 1.00 x 106 m/s enters a region in space where a magnetic fi eld givenby B = (–0.500 T) z exists. The velocity vector of the proton is at an angle θ = 60.0° with respect tothe positive z-axis. a) Analyze the motion of the proton and describe its trajectory (in qualitativeterms only). b) Calculate the radius, r, of the trajectory projected onto a plane perpendicular to themagnetic fi eld (in the xy-plane). c) Calculate the period, T, and frequency, f, of the motion in thatplane. d) Calculate the pitch of the motion (the distance traveled by the proton in the direction ofthe magnetic fi eld in 1 period).

An electron has an initial velocity of (10.0j + 14.5k) km/s and a constant acceleration of (1.70×1012 m/s2)i in a region in which uniform electric and magnetic fields are present. If B =(400μT)i, find the electric field E. Ex? Ey? Ez?

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^)