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  • An electron (mass $9 \times 10^{-31} \mathrm{~kg}$ and charge $1.6 \times 10^{-19} \mathrm{C}$ ) moving with speed $\mathrm{c} / 100$ ( $\mathrm{c}=$ speed of light) is injected into a magnetic field $\vec{B}$ of magnitude $9 \times 10^{-4}$ T

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

A) $\vec{E}$ is parallel to $\vec{B}$ and its magnitude is $27 \times 10^4 \mathrm{~V} \mathrm{~m}^{-1}$

B) $\vec{E}$ is perpendicular to $\vec{B}$ and its magnitude is $27 \times 10^4 \mathrm{Vm}^{-1}$

C) $\vec{E}$ is perpendicular to $\vec{B}$ and its magnitude is $27 \times 10^2 \mathrm{~V} \mathrm{~m}^{-1}$

D) $\vec{E}$ is parallel to $\vec{B}$ and its magnitude is $27 \times 10^2 \mathrm{~V} \mathrm{~m}^{-1}$

Answers (1)

best_answer

Solution:
For no deflection, the electric force must cancel the magnetic force:

$$
q E=q v B \rightarrow E=v B
$$


Given $v=c / 100=3 \times 10^6 \mathrm{~m} / \mathrm{s}$ and $\mathrm{B}=9 \times 10^{-2} \mathrm{~T}$,

$$
\mathrm{E}=\left(3 \times 10^5\right) \times\left(9 \times 10^{-2}\right)=27 \times 10^4 \mathrm{~V} / \mathrm{m}
$$


Electric field must be perpendicular to B and in opposite direction to magnetic force.

 

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Dimpy

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