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 Post subject: Physics: Moving Charges' Magnetic FieldsPosted: Fri, 9 Mar 2007 01:01:34 UTC
 S.O.S. Oldtimer

Joined: Tue, 20 May 2003 20:19:56 UTC
Posts: 225
Positive point charges q_1= 6.90 microC and q_2= 2.90 microC are moving relative to an observer at point P as shown in the figure. The distance from the observer to either charge is originally d = 0.190 m.The two charges are at the locations shown in the figure. Charge q_1 is moving at a speed of v_1 = 3.20Ã—10^6 m/s and q_2 is moving at a speed of v_2 = 9.40Ã—10^6 m/s.

a. What is the magnitude of the net magnetic field they produce at point P ?
b. What is the direction of the net magnetic field they produce at point P ?

I think I mixed up my signs for the cross product of v and r. Are the two fields in opposite directions? Any advice on those parts are appreciated.

B_1 = (mu_0/4*pi)*[(q*v vector*r vector)/(r^3)]

v vector = 3.20*10^6 m/s*i
r vector = (-0.190 m)*j

v vector*r vector = (3.20*10^6 m/s)i*(-0.190 m)j = -60800 m^2/s*k

B_1 = [(1*10^-7 T*m/A)*(6.90*10^-6 C)(-60800 m^2/s*k)]/[0.190 m]^3

= (-6.12*10^-5 T)*k ??

B_2 = (mu_0/4*pi)*[(q*v vector*r vector)/(r^3)]

v vector = -9.40*10^6 m/s*i
r = 0.190 m*j

v vector*r vector = (-9.40*10^6 m/s*i)*(0.190m*j) = -1786000 m^2/s

B_2 = [(1*10^-7 T*m/A)*(2.90*10^-6 C)(-1786000 m^2/s*k)]/[0.190 m]^3
= (-7.55*0^-5 T)*k ??

B_total = (-6.12*10^-5 T)*k - (-7.55*0^-5 T)*k = -(1.37*10^-4 T)*k out of the page?? (negative z-axis?)

_________________
Gracias for your time, patience, and input.

"Which is a kind of integrity, if you look on every exit being an entrance somewhere else."
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 Post subject: Posted: Fri, 9 Mar 2007 01:52:49 UTC
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The B-field contributions from each moving charge are pointing out towards you...

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