Question

An electron moves north (up) at a velocity of 3.8 ´ 105 m/s and has a magnetic force of 8.9 ´ 10–18 N west (left) exerted on it. What is the magnitude of the magnetic field?

Answer 1

`B = 1.464*10^(-4)T` or
`146.4\mu T`.

Step-by-step explanation:

The right hand rule tells us that to produce a magnetic force directed towards west, the magnetic field must point straight up from the plane, and therefore
`v` and
`B` are perpendicular to each other and hence

`F = q (v* B) = qvB`

Solving for
`B` we get

`B = (F)/(qv)`

putting in
`F = 8.9*10^(-18)N`,
`q = 1.6*10^(-19)C`, and
`3.8*10^(5)m/s` we get

`B = (8.9*10^(-18)N)/((1.6*10^(-19)C)(3.810^(5)m/s))`

`\boxed{B = 1.464*10^(-4)T.}`

which is also 146.4 micro Teslas.