I am stuck on this question:

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

I am stuck on this question:

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?

anonymous · Answer

Since $\vec F = q(\vec v \times \vec B)$
so, $\vec F = -evB sin \theta \hat n $
where $\theta$ is angle between the direction of travel and the Earth's magnetic and $\hat n$ is a unit vector perpendicular to the plane of $\vec v \ and \ \vec B$.
Put the given values and solve for $\theta$.

anonymous · Answer

thanks!

anonymous · Answer

is there a set value for e/q? (the value measured in Coulombs)

anonymous · Answer

I don't understand your question.. :(

anonymous · Answer

Oh sorry. In 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 thr angle (sin theta)
but I don't know what to put in place of q

anonymous · Answer

q is the charge of beta particle which is e = 1.6 * 10^-19