Q109 / 180 · AIIMS 2019 (26 May) · Physics

A proton of velocity $(3 \hat{i}+2 \hat{j}) \mathrm{m} \mathrm{s}^{-1}$ enters a field of magnetic induction $(2 \hat{j}+3 \hat{k})$ tesla. The acceleration produced in the proton in $\mathrm{m} \mathrm{s}^{-2}$ is (Specific charge of proton $=0.96 \times 10^8 \mathrm{C} \mathrm{kg}^{-1}$ )

  • A
    $2.8 \times 10^8(2 \hat{i}-3 \hat{j})$
  • Correct answer
    B
    $2.88 \times 10^8(2 \hat{i}-3 \hat{j}+2 \hat{k})$
  • C
    $2.8 \times 10^8(2 \hat{i}+3 \hat{k})$
  • D
    $2.88 \times 10^8(\hat{i}-3 \hat{j}+2 \hat{k})$

Solution

Here, $\vec{v}=3 \hat{i}+2 \hat{j} \mathrm{~m} \mathrm{~s}^{-1}$ $\vec{B}=2 \hat{j}+3 \hat{k} \mathrm{~T}$ Specific charge of proton $=\frac{e}{m}=0.96 \times 10^8 \mathrm{C} \mathrm{kg}^{-1}$ Force on a proton in a uniform magnetic field is $\begin{aligned} & \vec{F}=e(\vec{v} \times \vec{B}) \\ & =e[(3 \hat{i}+2 \hat{j}) \times(2 \hat{j}+3 \hat{k})] \\ & =e(6 \hat{k}-9 \hat{j}+6 \hat{i})=e(6 \hat{i}-9 \hat{j}+6 \hat{k}) \mathrm{N}\end{aligned}$ Acceleration of the proton, $\begin{aligned} & \vec{a}=\frac{\vec{F}}{m}=\frac{e(6 \hat{i}-9 \hat{j}+6 \hat{k})}{m} \\ & =0.96 \times 10^8(6 \hat{i}-9 \hat{j}+6 \hat{k}) \mathrm{m} \mathrm{s}^{-2} \\ & =2.88 \times 10^8(2 \hat{i}-3 \hat{j}+2 \hat{k}) \mathrm{m} \mathrm{s}^{-2}\end{aligned}$

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

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