Question

A particle of charge -3 C is momentarily located at the position (4,2,7) [all distances in meters]. The particle's velocity is 450i^+150j^+500k^ [in m/s] What is the magnetic field vector due to the particle at the position (8,6,7)? b.) At that same instant, a second particle of charge 11 C is momentarily located at the origin, moving along the positive x-axis with a speed of 2000 m/s. What is the total magnetic field at the target position from the previous problem?

Answer 1

Magnetic field due to a moving charge is given by

B = (/4)*q(v x r)/r³ {cross product of v and r}

we know, (/4) = 10^-7

r = (8i + 6j + 7k)-(4i + 2j + 7k) = 4i + 4j

r (magnitude) = 4*sqrt(2) = 5.656 m

v x r = (450i + 150j + 500k) x (4i + 4j)

= (-2000i + 2000j + 1200k)

putting all the values

B = 10^-7*(-3* (-2000i + 2000j + 1200k))/(5.656³)

B₁ = +3.31*10^-6i - 3.31*10^-6j - 1.988*10^-6k

b) due to the second charge

r = (8i + 6j + 7k)-(0i + 0j + 0k)

r (magnitude) = 4*sqrt(2) = 12.2 m

v x r = (450i + 150j + 500k) x (8i + 6j + 7k)

= (-1950i +850j + 1500k)

putting all the values

B = 10^-7*(11* (-1950i +850j + 1500k))/(12.2³)

B₂ = -14.3*10^-6i + 6.25*10^-6j + 11.03*10^-6k

total field

B = B₁ + B₂ = (-11i -2.94j +9.94k)*10^-6 T