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?

Respuesta :

Answer:

[tex]B = 1.464*10^{-4}T[/tex] or [tex]146.4\mu T[/tex].

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 [tex]v[/tex] and [tex]B[/tex] are perpendicular to each other and hence

[tex]F = q (v\times B) = qvB[/tex]

Solving for [tex]B[/tex] we get

[tex]B = \dfrac{F}{qv}[/tex]

putting in [tex]F = 8.9*10^{-18}N[/tex], [tex]q = 1.6*10^{-19}C[/tex], and [tex]3.8*10^{5}m/s[/tex] we get

[tex]B = \dfrac{8.9*10^{-18}N}{(1.6*10^{-19}C)(3.810^{5}m/s)}[/tex]

[tex]\boxed{B = 1.464*10^{-4}T.}[/tex]

which is also 146.4 micro Teslas.