A 6.60 (-)microC moves through a region of space where an electric field of magn

Question

A 6.60 (-)microC moves through a region of space where an electric field of magnitude 1300 n/c points in the positive x direction, and a magnetic field of magnitude 1.01T points in the positive z direction. If the net force acting on the particle is 6.22x10^-3N in the positive x direction, find the components of the particle's velocity. Assume the particle's velocity is in the x-y plane. v(x), v(y), v(z)= ?,?,? m/s

Explanation / Answer

First, you write your variables: Q = 6.60 x 10^-6 C E = 1300 N/C --->+x direction B = 1.01 T ------>+z direction Fnet = 6.22x 10^-3 N ----->+x direction Fnet = Fe - Fb =qE - qvB Solve for v so v = (1/B)(E - (Fnet/q)) once you get the value for velocity. You need to know which direction it's going. You know +x direction for E, +z direction for B and +x for Fnet. Do the right hand rule where: your right thumb goes toward the Fnet, then your index finger points to B (z direction) Then curl your middle,ring, and pink 90 angle. This shows where v is going which is -y direction. Answer: vx, vy,vz = 0, -(number you got for v), 0 make sure you put (-) because v is going - y direction

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