Question

An electron moves through a uniform electric field E = (2.60î + 5.90ĵ) V/m and a uniform magnetic field B = 0.400k T. Determine the acceleration of the electron when it has a velocity v = 6.0î m/s. (Give each component in m/s2.)

Answer 1

Answer:

The answer is "$1.75 \cdot (2.60 \hat{i} + 3.5 \hat{j}) \times 10^{11} \ \ \frac{m}{s^2}$ ".

Explanation:

Formula of acceleration =  

$\frac{F_e}{m_e} =\frac{-e(\underset{E}{\rightarrow} + \underset{V}{\rightarrow} \times \underset{B}{\rightarrow})}{m_e}$

values:

$\underset{E}{\rightarrow} = (2.60 \hat{i} + 5.90 \hat{j}) \frac{V}{m} \\\\\underset{B}{\rightarrow} = 0.400 k \ T \\\\\underset{V}{\rightarrow} = 6.0 \hat {i} \ \ \frac{m}{s} \\\\\ apply \ value \ in \ above \ formula: \\\\ \frac{F_e}{m_e} =\frac{e}{m_e} \cdot (2.60 \hat{i} + 5.90 \hat{j}+6.0 \hat {i} \times 0.4\hat {k} ) \\\\ \frac{F_e}{m_e} =\frac{e}{m_e} \cdot (2.60 \hat{i} + 5.90 \hat{j}- 2.4 \hat {j}) \\\\\therefore \frac{e}{m_e} = \frac{1.6 \times 10^{-19}}{9.1 \times 10^{-21}}$

$\frac{F_e}{m_e} = \frac{1.6 \times 10^{-19}}{9.1 \times 10^{-21}} \cdot (2.60 \hat{i} + 5.90 \hat{j}- 2.4 \hat {j}) \\\\\frac{F_e}{m_e} = \frac{1.6 \times 10^{-19}}{9.1 \times 10^{-21}} \cdot (2.60 \hat{i} + 3.5 \hat{j}) \\\\\frac{F_e}{m_e} = 1.75 \cdot (2.60 \hat{i} + 3.5 \hat{j}) \times 10^{11} \ \ \frac{m}{s^2}$