Exercise 28.6 Positive point charges q = 8.00 ?C and q?= 4.00 ?C are moving rela

Question

Exercise 28.6 Positive point charges q = 8.00 ?C and q?= 4.00 ?C are moving relative to an observer at point P, as shown in the figure (Figure 1) . The distance d is 0.130 m , v = 4.60×106 m/s , and v?= 9.20×106 m/s .

Part A

When the two charges are at the locations shown in the figure, what is the magnitude of the net magnetic field they produce at point P?

Part C

What is the magnitude of the electric force that each charge exerts on the other?

Part E

What is the magnitude of the magnetic force that each charge exerts on the other?

Part G

What is the ratio of the magnitude of the electric force to the magnitude of the magnetic force?

Explanation / Answer

Given that

The charge is q = 8.00 C =8*10-6C

The charge q= 4.00 C=4*10-6C

The distance between the each point charge d=0.130 m

The velocity of the first charge is v = 4.60×106 m/s

The velocity of the other charge is v= 9.20×106 m/s

a)

The magnitue of the net magnetic field at the point P is given by

Btotal =B+B'

=(uo/4pi)[qv/d2+q'v'/d2]

=10-7/(0.130)2[(8*10-6C)(4.60×106 m/s )+(4*10-6C)(9.20×106 m/s )]=4355.02*10-7T =4.355*10-4T

c)

The magnitude of electric force exerted on each other is

Fc =kqq'/(2d)2 =9*109*(8*10-6C)(4*10-6C)/2*(0.130)2 =8.52N

E)

The magnitude of magnetic force is given by FB =(uo/4pi)[qq'vv'/(2d)2]

=10-7[1354.24/2*(0.130)2]=4.006*10-3T

G)

The ratio of the magnitude of the electric force to the magnitude of the magnetic force?

FE/FB =8.52/4.006*10-3T =2126

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.