Question

How would you determine the magnitude of the magnetic field and the speed of `q_2`?

A charge `q_1 = 20.0 µC` moves with a speed of `4.00 ✕ 10^3` `m`/`s` perpendicular to a uniform magnetic field. The charge experiences a magnetic force of `7.46 ✕ 10^-3 N`. A second charge `q_2 = 7.00 µC` travels at an angle of `40.0°` with respect to the same magnetic field and experiences a `1.90 ✕ 10^-3 N` force.

Answer 1

`B=0.09325T`
`v=4.5times10^3 ms^-1`

Explanation:

The force, `F` on a charged particle, `q`, moving perpendicular to a magnetic field, `B` is given by:
`F=Bqv`
Where v= velocity of the particle.

So for `q_1` we have:
`7.46times 10^-3=B*(20times 10^-6)*(4times10^3`)
`B=0.09325T`

If moving at an angle, `theta` with respect to the field then the equation needs to be modified as:
`F=Bqvsintheta`

So for `q_2`

`1.9times 10^-3=0.09325*(7times10^-6).v.sin40`
`v=4528 ms^-1 =4.5times10^3 ms^-1`