Question

A proton is projected with velocity $\vec{v}=2 \hat{i}$ in a region where magnetic field $\vec{B}=(\hat{i}+3 \hat{j}+4 \hat{k}) \mu T$ and electric field $\vec{E}=10 \hat{i} \mu V m^{-1}$. Then find out the net acceleration of proton:

• A. 1400 $ms^{-2}$
• B. 700 $ms^{-2}$
• C. 1000 $ms^{-2}$
• D. 800 $ms^{-2}$

Answer 1

Answer: (A)

$\vec{F}=Q \vec{E}+Q(\vec{V} \times \vec{B})$
$\vec{F}=1.6 \times 10^{-19} \times 10 \hat{i} \times 10^{-6}+1.6 \times 10^{-19}[(2 \hat{i}) \times(\hat{i}+3 \hat{j}+4 \hat{k})] \times 10^{-6}$
$\vec{F}=1.6 \times 10^{-19}[10 \hat{i}+6 \hat{k}-8 \hat{j}] \times 10^{-6}$
$\vec{F}=1.6 \times 10^{-19}[10 \hat{i}-8 \hat{j}+6 \hat{k}] \times 10^{-6} N$
$\vec{a}=1.6 \times 10^{-19}[10 \hat{i}-8 \hat{j}+6 \hat{k}] \times 10^{-6} / 1.6 \times 10^{-27} m / s ^{2}$
$1400 m / s ^{2}$