Positive point charges q = +7.75 C and q, = +2.50 C are moving relative to an ob
ID: 1581950 • Letter: P
Question
Positive point charges q = +7.75 C and q, = +2.50 C are moving relative to an observer at point P, as shown in the following figure. The distance d is 0.200 m, v = 4.15 x 106 m/s, and v' 8.70 x 106 m/s. (a) When the two charges are at the locations shown in the figure, what are the magnitude and direction of the net magnetic field they produce at point P? magnitude direction into the page (b) What are the magnitude and direction of the electric and magnetic forces that each charge exerts on the other on charge electric force magnitude direction upward magnetic force magnitude direction upward on charge q electric force magnitude direction downward magnetic force magnitude direction donward What is the ratio of the magnitude of the electric force to the magnitude of the magnetic force? 8 (c) If the direction of ' is reversed, so both charges are moving in the same direction, what are the magnitude and direction of the magnetic forces that the two charges exert on each other? on charge q: magnetic force magnitude direction donward on charge q: magnetic force magnitude direction dExplanation / Answer
(a) Magnetic field due to charge q
B1 = (uo /4Pi)*(qv/r2) = 10-7*[(7.75*10-6*4.15*106) /(0.2)2] = 8.04*10-5 T
Magnetic fied due to charge q'
B2 = (uo /4Pi)*(q'v'/r2) = 10-7*[(2.5*10-6*8.7*106) /(0.2)2] = 5.43*10-5 T
Since both magnetic field are subjected into the page , therefore
net magnetic field , B = B1 + B2 = 8.04*10-5 + 5.43*10-5 = 13.47*10-5 T
(b) on charge q
Force (Fe)= kqq' /r2 = 9*109*(7.75*10-6*2.5*10-6) /(2*0.2)2 = 1.089 N
magnetic force on q
Fm = qV*B
B due to charge q'
B = (uo /4Pi)*(q'v'/r2) here r will be = 2d becasue we are finding magnetic field at charge q .
= 10-7*[(2.5*10-6*8.7*106) /(2*0.2)2] = 1.35*10-5 T
Fm = (7.75*10-6)*(4.15*106)*(1.35*10-5) = 4.342*10-4 N
Forces on the charge q' will be same in magnitude but direction are different which you have marked correct.
(c) Ratio of forces
FC /FM = 1.089 / 4.342*10-4 = 2.51*103
(d) Magnetic force on charge q
F = qvB
Where B = (uo/ 4Pi)(q'V' /r2)
F = qv*[(uo/ 4Pi)(q'V' /r2)]
which is same what we calculated earlier that means the magnitude will remain same but direction will change.
Magnetic force will remain same for both charge = 4.342*10-4 N