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

at right angles to each other

t = 1/400 h   ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.