Question

An electron moves through a uniform electric field E = (2.00î + 5.40ĵ) V/m and a uniform magnetic field B = 0.400k T. Determine the acceleration of the electron when it has a velocity v = 8.0î m/s. (Give each component in m/s2.)

Answer 1

Answer:$a=1.75\times 10^{21}\left ( 2\hat{i}5.08\hat{j}\right )$

Explanation:

Given

Electric Field $\vec{E}=2\hat{i}+5.4\hat{j}$

$\vec{B}=0.4\hat{k}\ T$

velocity $\vec{v}=8\hat{i} m/s$

mass of electron $m=9.1\times 10^{-31} kg$

Force on a charge Particle moving in Magnetic Field

$F=e\left [ \vec{E}+\left ( \vec{v}\times \vec{B}\right )\right ]$

$a=\frac{e\left [ \vec{E}+\left ( \vec{v}\times \vec{B}\right )\right ]}{m}$

$a=\frac{1.6\times 10^{-19}\left [ 2\hat{i}+5.4\hat{j}+\left ( 8\hat{i}\times 0.4\hat{k}\right )\right ]}{9.1\times 10^{-31}}$  

$a=1.75\times 10^{21}\left ( 2\hat{i}+5.08\hat{j}\right )\ m/s^2$