A negative charge q = 3.40×106 C is located at the origin and has velocity V =(7

Question

A negative charge q = 3.40×106 C is located at the origin and has velocity V =(7.50×104m/s)^+((4.90)×104m/s)j^.At this instant what is the magnetic field produced by this charge at the point x = 0.250 m , y = -0.350 m , z = 0?

I know the relevant equation is m0/4pi*qVxr/r^2, I understand that q = -3.4*10^-6 C, and that the velocity is broken down in i^, j^, k^ notation I am getting confused on how to calculate r and how to apply it to both the denominator and numerator. It is also a little murky to me on how to multiply the charge through the equation.

Explanation / Answer

q = - 3.4 x 10^-6 C

v = 7.5 x 10^4 i - 4.9 x 10^4 j

r = 0.25 i - 0.350 j

B = (uo / 4 pi ) q (v x r )/ r3.

B = 10^-7 * ( - 3.4 * 10^-6 ) * (7.5 x 10^4 i - 4.9 x 10^4 j) x (0.25 i - 0.350 j) / (sqrt(0.25^2 + 0.35^2))^3

B = 10^-7 * ( - 3.4 * 10^-6 ) * ( - 26250 k + 12250 k) / (sqrt(0.25^2 + 0.35^2))^3

B = 10^-7 * ( - 3.4 * 10^-6 ) * ( - 14000k) /0.0796 = 5.98x 10^-8 T k

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