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)/r^{3} {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))/(5.656^{3})

B_{1} = +3.31*10^{-6}i - 3.31*10^{-6}j -
1.988*10^{-6}k

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))/(12.2^{3})

B_{2} = -14.3*10^{-6}i + 6.25*10^{-6}j +
11.03*10^{-6}k

total field

B = B_{1} + B_{2} = (-11i -2.94j
+9.94k)*10^{-6} T

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