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

Two complex values are z1=8 + 8i, z2=15 + 7 i. z1∗ and z2∗ are the complex conjugate values. Any complex value can be expessed in the form of a+bi=reiθ. Find r and θ for z1z2∗. Find r and θ for z1/z2∗? Find r and θ for (z1−z2)∗/z1+z2∗. Find r and θ for (z1−z2)∗/z1z2∗ Please explain all steps, Thank you

An ac series circuit consists of a voltage source of frequency 60 Hz and voltage amplitude V, a 505-Ω resistor, and a capacitor of capacitance 7.2 μF. What must be the source voltage amplitude V for the average electrical power consumed in the resistor to be 236 W? There is no inductance in the circuit.

An L−R−C series circuit has R= 280 Ω . At the frequency of the source, the inductor has reactance XLL= 905 Ω and the capacitor has reactance XC= 485 Ω . The amplitude of the voltage across the inductor is 445 V . What is the amplitude of the voltage across the resistor and the capacitor? What is the voltage amplitude of the source? What is the rate at which the source is delivering electrical energy to the circuit?

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^)