Positive point charges q = 7.00 ? C and q ?= 4.00 ? C are moving relative to an

Question

Positive point charges q = 7.00 ?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 0.110 m , v = 4.60×106 m/s , and v?= 9.20×106 m/s .

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?

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

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

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

Explanation / Answer

A)

magnetic field due to moving point charge = B = uo*q*v*sintheta/r

B = (uo*q*v*sintheta)/4*pi*d

B' = (uo*q'*v'*sintheta)/4*pid

Bnet = B + B' = (uo*(qv + q'v') *sintheta)/4*pi*d

here theta = 90

Bnet = (4*pi*10^-7*((7*10^-6*4.6*10^6)+(4*10^-6*9.2*10^6))*sin90)/(4*pi*0.11)

Bnet = 6.27*10^-5 T

++++++

(B)

Fe = k*q*q/(2d)^2

Fe = (9*10^9*7*10^-6*4*10^-6)/(4*0.11^2) = 5.21 N

-----------------

(C)

on q

B' = (uo*q'*v'*sintheta)/4*pi*2d

Fb = B'*q*v= (uo*q'*v'*sintheta*v*q)/4*pi*2d

on q'

B = (uo*q*v*sintheta)/4*pi*2d

Fb' = B*q'*v' = (uo*q'*v'*sintheta*v*q)/4*pi*2d

Fb = Fb' = (4*pi*10^-7*7*10^-6*4*10^-6*4.6*10^6*9.2*10^6)/(4*pi*2*0.11)

Fb = Fb' = 0.000538 N ,<-----answer

part(D)

ratio = Fe/Fb = 5.21/0.000538 = 9684.01

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