Question

When a proton has a velocity v=2i+3j×10^6 m per s, it experiences a force F=-1.28×10^-13k N.when its velocity is along the z axis then it experience a force along the x axis what is the magnetic field

Answer 1

Answer:

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 }