Question

An alpha particle (proton) is moving with a velocity of 820.0 m/s through a magnetic field which has a field intensity of 1.20 Tesla. Assuming that this particle is moving at right angles to the field what will be the magnitude of the resultant magnetic force on it?

Answer 1

The magnitude of the resultant magnetic force on the alpha particle is 3.84x10^-13 Newtons.

Step-by-step explanation:

The magnitude of the resultant magnetic force on the alpha particle can be calculated using the formula:

F = qvB

Where F is the force, q is the charge of the particle (in this case, the charge of an alpha particle is twice the charge of a proton), v is the velocity of the particle, and B is the magnetic field intensity.

Plugging in the values, we have:

F = (2*q)*(v)*(B)

F = (2*(1.6x10^-19 C))*(820.0 m/s)*(1.20 Tesla)

F = 3.84x10^-13 N

The magnitude of the resultant magnetic force on the alpha particle is 3.84x10^-13 Newtons.