A particle of charge -16 x 10^-18 C moving with velocity 10 ms^-1 along the x-axis enters a region
Answers
Mag v = sqrt(13) x10^6 m/s
F = 1.28x10^-13 N
q = 1.6 x 10^-19 C
B = F / qv = 1.28x10^-13 N /(1.6 x 10^-19 C)(3.6 x 10^3)
B = 0.222 x 10^3Tesla
So B = -0.222k x 10^3 Tesla = -222 Tesla in k direction
{You can use:
F = qv x B, (ignore the cross product though, just do a normal multiplication) to find the field's strength, using the information from the first part of the question (rearrange it for B of course).
For the second part, you can just use the right-hand rule. Going back to that formula:
F = qv x B
You point your fingers in the direction of v, then curl them to B and your thumb points to F.
You don't know where B is, but you know v and F, so just point your fingers towards v, and your thumb towards F to find B (it will be the direction your palm points in).
B is in the -y direction.
EDIT: Just a reminder for the unit vectors,
i corresponds to the +x-axis
j to the +y-axis
k to the +z-axis
If the velocity were along the z-axis and the magnetic field along the -z-axis, you wouldn't get any force at all (cross-product of parallel vectors is zero }