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

Correct answer: A. 1400 ~m ~s^ -2 aligned & F =q E +q( v × B ) & F =1.6 × 10^ -19 × 10 i × 10^ -6 +1.6 × 10^ -19 & [(2 i ) ×( i +3 j +4 k )] × 10^ -6 & F =1.6 × 10^ -19 [10 i +6 k -8 j ] × 10^ -6 N & a =1.6 × 10^ -19 [10 i -8 j +6 k ] × 10^ -6 / 1.6 × 10^ -27 m s^ -2 & a=| a |=100 200 =1.41 × 1000 1400 m s^ -2 aligned

AIIMS 201925 May

Exam: AIIMS
Year: 2019
Date: 25 May
Subject: Physics
Chapter: Magnetic Effects of Current

AIIMS 2019 · 25 May · Physics · Magnetic Effects of Current

A proton is projected with velocity v =2 i in a region where...

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

Options

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

Correct answer

A. $1400 \mathrm{~m} \mathrm{~s}^{-2}$

Step-by-step 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}

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