Magnetic Field Intensity and Point Charges

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SUMMARY

The discussion focuses on calculating the net magnetic field intensity produced by two moving point charges, q_1 (6.90 µC) and q_2 (2.90 µC), at a point P located 0.190 m away. Charge q_1 moves at 3.20×10^6 m/s, while q_2 moves at 9.40×10^6 m/s. The magnetic fields B_1 and B_2 were computed using the formula B = (μ₀/4π) * (q * v × r) / r³, resulting in B_total = -(1.37×10^-4 T) k, indicating the direction is out of the page along the negative z-axis. The participant seeks confirmation on the correctness of their calculations and signs.

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  • Understanding of magnetic field calculations using the Biot-Savart Law
  • Familiarity with vector cross products
  • Knowledge of the properties of point charges and their motion
  • Basic principles of electromagnetism, specifically magnetic fields due to moving charges
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  • Study vector cross product operations in physics
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Students studying electromagnetism, physics educators, and anyone involved in advanced physics problem-solving related to magnetic fields and point charges.

Soaring Crane
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Homework Statement



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 = 3.20×10^6 m/s and q_2 is moving at a speed of v_2 = 9.40×10^6 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 ?

yf_Figure_28_29.jpg


Homework Equations



See below.

The Attempt at a Solution


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?)

Thanks.
 
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Can anyone please check over my workings, particularly my signs?

Thank you very much.
 

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