A beta particle (fast moving electron) is released by a nuclear decay and starts travelling at 1.5x10^6 m/s [NΘE] in the Earth's magnetic field which has a value of 3.0x10^-5 T. If the force experienced by the beta particle is 4.34x10^-18 N, determine the angle between the direction of travel and the Earth's magnetic field. What is the direction of the force acting on the particle?

Question

Using the formula f=qvBsin(theta):
the f value would be 4.34x10^-18 N
the B value would be 3.0x10^10^-5 T
the v value would be 1.5x10^6 m/s
the question asks for the angle (sin theta)
but I don't know what to put in place of q

Answer 1

q is the electron charge

-1.6*10^-19 Coulombs

You are supposed to know that, The charge on a proton is equal and opposite in sign.