Question

An electron moves at 2.80 ×106 m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.60 ×10−2 T. Part A What is the largest possible magnitude of the acceleration of the electron due to the magnetic field? Express your answer with the appropriate units. aa = nothing

Answer 1

Answer:

$a = 3.73*10^{16}m/s^2$

Explanation:

The magnetic force is given by the equation,

$\vec{F}_{magnetic} = q (\vec{v}X\vec{B})$

$\vec{F}_{magnetic} = qvBsin\theta$

Where $\theta$ is the angle between velocity vector and the magnetic field,

That angle is 90°.

We know as well that

F=ma, replacing the mass and the acceleration in our previous equation we have

$ma = qvBsin\theta$

$a= \frac{qvBsin\theta}{m}$

Our values are,

$q= 1.6*10^{-19}c$

$v=2.8*10^6m/s$

$B=7.6*10^{-2}T$

$m=9.11*10^{-31}kg$

Substituting,

$a = \frac{(1.6*10^{-19})(2.8*10^6)(7.6*10^{-2})}{9.11*10^{-31}}$

$a = 3.73*10^{16}m/s^2$