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A particle with charge 6.00 nC is moving in a uniform magnetic field B =( 1.30 T )k^. The magnetic force on the particle is measured to beF =( 3.10×107 N )i^+( 7.60×107 N )j^.

Part B

Calculate thex-component of the velocity of the particle.

Part C

Calculate they-component of the velocity of the particle.

Part D

Calculate the scalar product v x F

Part E

What is the angle between v  and F ? Give your answer in degrees.

Explanation / Answer

B) as we know that F=q*(V×B) = q*(-1.3*Vy i cap +1.3*Vx j cap)

On comparing with given force

7.6*10^-7=q*(1.3*Vx)

Vx= 7.6*10^-7/(1.3*(-6*10^-9))= -97.44

C)Vy=-3.1*10^-7/(1.3*(-6*10^-9))=-39.74

D) V×F=k cap(-97.44*7.6*10^-7-39.74*3.1*10^-7) = -863.71*10^-7 k cap

E) V×F =|V|*|F|*sin thetha

|V|=(-97.44)^2+(-39.74)^2= 105.23

|F|=(-3.1*20^-7)^2+(7.6*10^-7)^2=8.2*10^-7

Sin thetha = 863.71*10^-7/(105.23*8.2*19^-7) =1

Thetha= 90°

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