Q104 / 180 · AIIMS 2019 (25 May) · Physics

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

  • Correct answer
    A
    $1400 \mathrm{~m} \mathrm{~s}^{-2}$
  • B
    $700 \mathrm{~m} \mathrm{~s}^{-2}$
  • C
    $1000 \mathrm{~m} \mathrm{~s}^{-2}$
  • D
    $800 \mathrm{~m} \mathrm{~s}^{-2}$

Solution

$\begin{aligned} & \text { } \vec{F}=q \vec{E}+q(\vec{v} \times \vec{B}) \\ & \vec{F}=1.6 \times 10^{-19} \times 10 \hat{i} \times 10^{-6}+1.6 \times 10^{-19} \\ & \quad[(2 \hat{i}) \times(\hat{i}+3 \hat{j}+4 \hat{k})] \times 10^{-6} \\ & \vec{F}=1.6 \times 10^{-19}[10 \hat{i}+6 \hat{k}-8 \hat{j}] \times 10^{-6} \mathrm{~N} \\ & \vec{a}=1.6 \times 10^{-19}[10 \hat{i}-8 \hat{j}+6 \hat{k}] \times 10^{-6} / 1.6 \times 10^{-27} \mathrm{~m} \mathrm{~s}^{-2} \\ & a=|\vec{a}|=100 \sqrt{200}=1.41 \times 1000 \approx 1400 \mathrm{~m} \mathrm{~s}^{-2}\end{aligned}$

Hence, the correct answer is option: 'A'.

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