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  • An Electron experiences a force  <img alt="(4 \hat{i}+3 \hat{\jmath}) \times 10^{-13} \mathrm{~N}" src="https://entrancecorner.oncodecogs.com/gif.latex?%284%20%5Chat%7Bi%7D&plus;3%20%5Chat

An Electron experiences a force  (4 \hat{i}+3 \hat{\jmath}) \times 10^{-13} \mathrm{~N} in a uniform magnetic field when its velocity is  2.5 \,\hat{k} \times 10^7 \mathrm{~m} / \mathrm{s}. When the velocity is redirected and becomes  (1.5 \hat{i}-2.0 \hat{j}) \times 10^7 \mathrm{~m} / \mathrm{s}, the magnetic force of the electron is zero. The magnetic field vector  \vec{B}  is:

 

Option: 1

-0.075\, \hat{i}+0.1\, \hat{j}


Option: 2

0.1 \, \hat{i}+0.075\, \hat{j}


Option: 3

0.075\, \hat{i}-0.1\, \hat{j}+ \hat{k}


Option: 4

0.075\, \hat{i}-0.1\, \hat{j}


Answers (1)

{\vec{B}} =Bx \hat{i}+B y \hat{\jmath}+Bz \hat{k} \\

      =(4.0 \hat{i}+3.0 \hat{\jmath}) \times 10^{-13} \\
      =-e\left(2.5 \hat{k}\times \left(Bx \hat{i}+By \hat{\jmath}+B z \hat{k}\right) \times 10^7\right) \\
      =(-2.5 e B x \hat{j}+2.5 e B y \hat{i}) \times 10^7 \\

\quad 10^{-13} \times 4 =2.5 \times 1.6 \times 10^{-19} B y \times 10^7 \\

              \quad B y =0.1 \\

             B x =-0.075

B =B x \hat{i}+B y \hat{j} \\

       =-0.075 \hat{i}+0.1 \hat{J}=-0.075 \hat{i}+0.1 \hat{j}
 

Posted by

Kshitij

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