A particle with positive charge q = 4.65 times 10^-18 C moves with a velocity v^

Question

A particle with positive charge q = 4.65 times 10^-18 C moves with a velocity v^rightarrow = (3i + 2j - k) m/s through a region where both a uniform magnetic Reid and a uniform electric field exist. (a)Calculate the total force on the moving particle, taking B^rightarrow = (4i + 5j + k) T and E^rightarrow = (2i - j - 5k) V/m. f_x = The force will be the sum of the force due to the electric field and the force due to the magnetic field. N F_y = N F_z = N (b) What angle does the force vector make with the positive x axis? degree

Explanation / Answer

(a) electric force= Fe= q E=4.65x10-18(2i-j-5k) N

magnetic force=q (VxB)=4.65x10-18 [(3i+2j-k)x(4i+5j+k)]= 4.65x10-18 (7i-7j+7k)N

total force F= Fe+ Fm=4.65x10-18 (2i-j-5k) +4.65x10-19 (7i-7j+7k)

F=4.65x10-18 (9i-8j+2k)

F= (4.185x10-17 i -3.72x10-17 j +9.3 x10-18k)N

Fx=4.185x10-17 N

Fy=-3.72x10-17 N

Fz=9.3x10-18 N

(b)magnitude of force F=sqrt[(4.185x10-17 )2+(-3.72x10-17)2 +(9.3 x10-18)2]=5.68x10-17 N

for angle with x axis

cos theta= Fx/F= 4.185x10-17/5.68x10-17

cos theta=0.737

theta=42.5 degrees

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