Question

Magnetic fields are produced by moving charges and exert
forces on moving charges. When a particle with charge q is moving
with velocity v
S
in a magnetic field B
S
, the force F
S
that the field exerts
on the particle is given by F
S
= qv
S
: B
S
. The SI units are as follows: For
charge it is the coulomb (C), for magnetic field it is tesla (T), for force
it is newton (N), and for velocity it is m>s. If q = -8.00 * 10-6 C, v
S
is 3.00 * 104 m>s in the +x-direction, and B
S
is 5.00 T in the -y-direction, what are the magnitude and direction of the force that the magnetic
field exerts on the charged particle?

Answer 1

F = q V cross B
q = -8 * 10^-6
V = 3 * 10^4 in x direction
B = 5 in - y direction

V cross B is in NEGATIVE z direction by right hand rule with positive x cross negative y
since q is also negative, the final result is in the PLUS z direction
so
+ 8*10^-6 * 3*10^4 * 5 Newtons