Question

A stream of doubly ionized particles (2 protons) moves at a velocity of 3.0 × 104 m/s perpendicularly to a magnetic field of 0.0900 T. What is the magnitude of the force on the particles? (answer in scientific notation, including sign)Equations F = BqvF = BILcharge on a proton = 1.6 x 10–19C  charge on an electron = –1.6 x 10–19C

Answer 1

The force on a charged particle on a magnetic field is given by:

$F=Bqv\sin\theta$

where B is the magnitude of the field, q is the charge of the particle, v is its velocity and θ is the angle between the field and the velocity of the particle.

In this case we have that:

• The magnitude of the field is 0.09 T.

,

• The charge of the particle is 3.2 x 10 –19C .

,

• The velocity is  3.0 × 10 4 m/s

,

• The angle between the field and the velocity is 90°, since they are perpendicular.

Plugging these we have:

$\begin{gathered} F=(0.09)(3.2\times10^{-19})(3\times10^4)\sin90 \\ F=8.64\times10^{-16}\text{ N} \end{gathered}$

Therefore, the force on the charge is:

$F=8.64\times10^{-16}\text{N}$