Question

a particle with a charge of 0.030 C experiences a magnetic force of 1.5 N while moving at right angles to a uniform magnetic field. If the speed of the charge is 620 m/s, what is the magnitude of the magnetc field the particle passes through?

Answer 1

The magnetic force experienced by a charged particle moving at right angle in a magnetic field is given by:
$F=qvB$
where
q is the charge
v is the speed of the particle
B is the intensity of the magnetic field

In our problem, q=0.030 C, v=620 m/s and F=1.5 N, therefore if we re-arrange the equation and we plug these data into it, we find the intensity of the magnetic field:
$B= \frac{F}{qv}= \frac{1.5 N}{(0.030 C)(620 m/s)}=0.08 T$