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Get Answers to all your Questions
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)
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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