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  • An electron enters an electric field having intensity <img alt="\vec{E}=3 \hat{j}+6 \hat{j}+2 \hat{k}\ V \cdot m^{-1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cvec%7BE%7D%3D3%20

An electron enters an electric field having intensity \vec{E}=3 \hat{j}+6 \hat{j}+2 \hat{k}\ V \cdot m^{-1} and a magnetic field having induction \vec{B}=2 \hat{i}+3 \hat{J}\ T with a velocity \vec{v}=\hat{2} i+3 \hat{j} \mathrm{~m} . s^{-1}. The magnitude of the force acting on the electrons in ( Given e =  1.6 \times 10^{-19 } e )

Option: 1

2.02 \times 10^{-18} \mathrm{~N}


Option: 2

5.16 \times 10^{-16} \mathrm{~N}


Option: 3

3.72 \times 10^{-17} \mathrm{~N}


Option: 4

None of the above


Answers (1)

best_answer

Force on the electron

\vec{F}=e(\vec{E}+\vec{v} \times \vec{B})

Now, \vec{V} \times \vec{B}=\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 0 \\ 2 & 3 & 0 \end{array}\right|=\overrightarrow{0}\therefore \vec{F}=1.6 \times 10 \times 10^{-19}(3 \hat{i}+6 \hat{j}+2 \hat{k}+\overrightarrow{0})

\therefore Magnitude of force = \text { 1. } 6 \times 10^{-19} \sqrt{3^2+6^2+2^2}

=1.12 \times 10^{-18} \mathrm{~N}

none of the above option correct.

Posted by

jitender.kumar

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