[20 points] A particle of charge of 2 × 10-⁹ [C] is moving with a velocity equal to 7 = ( 4 × 105)-(1 x 105) Ĵ +(2 × 10³

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[20 points] A particle of charge of 2 × 10-⁹ [C] is moving with a velocity equal to 7 = ( 4 × 105)-(1 x 105) Ĵ +(2 × 10³

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20 Points A Particle Of Charge Of 2 10 C Is Moving With A Velocity Equal To 7 4 105 1 X 105 J 2 10 1
20 Points A Particle Of Charge Of 2 10 C Is Moving With A Velocity Equal To 7 4 105 1 X 105 J 2 10 1 (29.37 KiB) Viewed 13 times
[20 points] A particle of charge of 2 × 10-⁹ [C] is moving with a velocity equal to 7 = ( 4 × 105)-(1 x 105) Ĵ +(2 × 10³) k [m/s] is in a region of magnetic field given by B=0.2 +0.1 -0.2 k [T]. A magnetic force will be applied on the particle. a) [5 points] Find the Fx, the x-component of F. Fx = [N] b) [5 points] Find the Fy, the y-component of F. Fy= [N] c) [5 points] Find the F₂, the z-component of F. F₂ = [N] d) [5 points] Find the magnitude of magnetic force F F = [N]
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