Question

What must be the charge on a particle if a force of 8.13N is applied when it travels at 2.61m/s through a magnetic field of 2.78T?0.892C1.12C0.115C8.66C

Answer 1

The magnitude of the magnetic force is given by:

`F=qvB\sin\theta`

where q is the charge of the particle, v is the velocity, B is the magnitude of the magnetic field and theta is the angle between the velocity and the magnetic field. Since we are not given any angle between the velocity and the field we will assume, for simplicity, that they are perpendicular, that is, we will assume the angle is 90°. We know that the force is 8.13 N, the velocity is 2.61 m/s and the field magnitude is 2.78 T; plugging these values we have:

`\begin{gathered} 8.13=(2.61)(2.78)(\sin90)q \\ q=(8.13)/((2.61)(2.78)(\sin90)) \\ q=1.12 \end{gathered}`

Therefore, the charge on the particle is 1.12 C