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Question

An electron (mass $9 \times 10^{- 31}\text{ }kg$ and charge $1.6 \times 10^{- 19}C$ ) moving with speed $c/100(c =$ speed of light) is injected into a magnetic field $\overrightarrow{B}$ of magnitude $9 \times 10^{- 4}\text{ }T$ perpendicular to its direction of motion. We wish to apply an uniform electric field $\overrightarrow{E}$ together with the magnetic field so that the electron does not deflect from its path. Then (speed of light $c = 3 \times 10^{8}{\text{ }ms}^{- 1}$ )

Options

A.$\overrightarrow{E}$ is perpendicular to $\overrightarrow{B}$ and its magnitude is $27 \times 10^{4}\text{ }V{\text{ }m}^{- 1}$
B.$\overrightarrow{E}$ is perpendicular to $\overrightarrow{B}$ and its magnitude is $27 \times 10^{2}\text{ }V{\text{ }m}^{- 1}$
C.$\overrightarrow{E}$ is parallel to $\overrightarrow{B}$ and its magnitude is $27 \times 10^{2}\text{ }V{\text{ }m}^{- 1}$
D.$\ \overrightarrow{E}$ is parallel to $\overrightarrow{B}$ and its magnitude is $27 \times 10^{4}\text{ }V{\text{ }m}^{- 1}$

Solution

For no deflection of electron, ${\overrightarrow{F}}_{B} = {\overrightarrow{F}}_{E}$

$$\begin{matrix} & \ - e(\overrightarrow{v} \times \overrightarrow{B}) = - e\overrightarrow{E} \\ & \ \Rightarrow \overrightarrow{E} = \overrightarrow{v} \times \overrightarrow{B} \Rightarrow \overrightarrow{E}\bot\overrightarrow{B} \\ & E = vB = \frac{c}{100} \times 9 \times 10^{- 4} \\ & \ = \frac{3 \times 10^{8}}{100} \times 9 \times 10^{- 4} \\ & \ = 27 \times 10^{2}\text{ }V{\text{ }m}^{- 1} \end{matrix}$$

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