The magnetic force on a charge.

Question

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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^) .

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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^) .

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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^) .

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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^) .

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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^)