Use F = qv (?) B.
Let By and Bz be the magnetic field components in the y-direction and z-direction, respectively.
Let Fy and Fz be the forces caused by By and Bz, respectively.
Since the velocity is normal to both axes, the angle between the vectors when the cross product is computed is 90(?) , and sin(90(?) ) = 1.
Fy = qv (?) By = qvBsin(90(?) )
= (34 (?) C)(74 m/s)(0.41 T)(1).
One tesla = one newton per ampere per meter, and one ampere is one coulomb per second, so rewrite this as
Fy = (34 (?) 10^(-6) C)(74 m/s)(0.41 N/(m(?) C/s)) = 1.0 (?) 10^(-3) N, or 1.0 mN.
Fz = qv (?) Bz = qvBsin(90(?) )
= (34 (?) C)(74 m/s)(0.86 T)(1)
= 2.1 (?) 10^(-3) N, or 2.1 mN.
a) The magnitude of the resultant is ?((1.0)(?) + (2.1)(?) ) = 2.3 (?) 10^(-3) N, or 2.3 mN.
b) Fz is the shorter leg of a right triangle; Fy is the longer leg. (The resultant vector is the hypotenuse.)
The angle ? from the z-axis to the resultant vector is tan?(?) (Fz/Fy)
= tan?(?) (2.1 mN / 1.0 mN)
= 64(?) (approximately)