The magnetic field in a region is given by B = (0.750 + 0.270) T. At some instant, a particle with charge q = 28.0 mC has velocity v = (35.6 + 107.3 + 44.5) m/s. What is the magnetic force exerted on the particle at that instant? (Express your answer in vector form.)

F = (-0.33642, 0.9345,- 1.984164) N

Explanation:

The magnetic force can be calculated as:

F = q*(vxB)

v = (vx, vy, vz) = (35.6, 107.3, 44.5) m/s

B = (Bx, By, Bz) = (0.750, 0.270, 0 ) T

where the cross product can be calculated as:

the cross product is:

vxB = ((vy*Bz - vz*By), (vz*Bx - vx*Bz), (vx*By - vy*Bx))

now, we can replace the values and get:

vxB = (107.3*0 - 44.5*0.270), (44.5*0.750 - 0), ( 35.6*0.270 - 0.750*107.3)) T*m/s

      = (-12.015, 33.375, -70.863) T*m/s

Then the magnetic force is:

F = (28.0)mC * (-12.015, 33.375, -70.863) T*m/s

remember that the unit of Teslas is in coulombs, so we must writ 28mC as

F = (0.028)*C* (-12.015, 33.375, -70.863) T*m/s

F = (-0.33642, 0.9345,- 1.984164) N

lexy is right because she has the labeled s

i think it’s a. this is because if they measured the mass of something with a mass of 150g, then their final answer would have to be approximately 150g. though all of their answers are almost 150, rosa and tran have the closest answers (or the exact answer) to the accurate one.