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AIIMS 201926 May
Exam
AIIMS
Year
2019
Date
26 May
Subject
Physics
Chapter
Magnetic Effects of Current

AIIMS 2019 · 26 May · Physics · Magnetic Effects of Current

A proton of velocity (3 i +2 j ) m s^ -1 enters a field of magnetic...

A proton of velocity (3i^+2j^)ms−1(3 \hat{i}+2 \hat{j}) \mathrm{m} \mathrm{s}^{-1} enters a field of magnetic induction (2j^+3k^)(2 \hat{j}+3 \hat{k}) tesla. The acceleration produced in the proton in ms−2\mathrm{m} \mathrm{s}^{-2} is (Specific charge of proton =0.96×108Ckg−1=0.96 \times 10^8 \mathrm{C} \mathrm{kg}^{-1} )

Options

  1. A2.8×108(2i^−3j^)2.8 \times 10^8(2 \hat{i}-3 \hat{j})
  2. B2.88×108(2i^−3j^+2k^)2.88 \times 10^8(2 \hat{i}-3 \hat{j}+2 \hat{k})
  3. C2.8×108(2i^+3k^)2.8 \times 10^8(2 \hat{i}+3 \hat{k})
  4. D2.88×108(i^−3j^+2k^)2.88 \times 10^8(\hat{i}-3 \hat{j}+2 \hat{k})

Correct answer

B. 2.88×108(2i^−3j^+2k^)2.88 \times 10^8(2 \hat{i}-3 \hat{j}+2 \hat{k})

Step-by-step solution

Here, v⃗=3i^+2j^ m s−1B⃗=2j^+3k^ TSpecificcharge\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 =em=\frac{e}{m}=0.96 \times 10^8 Ckg−1\mathrm{C} \mathrm{kg}^{-1} Force on a proton in a uniform magnetic field is F⃗=e(v⃗×B⃗)=e[(3i^+2j^)×(2j^+3k^)]=e(6k^−9j^+6i^)=e(6i^−9j^+6k^)N\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, a⃗=F⃗m=e(6i^−9j^+6k^)m=0.96×108(6i^−9j^+6k^)ms−2=2.88×108(2i^−3j^+2k^)ms−2\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}
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