An electron (mass kg and charge C) is moving with speed (= speed of light) is injected into a magnetic field of magnitude T perpendicular to its direction of motion. We wish to apply a uniform electric field together with the magnetic field so that th

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Solution

Q.
An electron (mass $9 \times 10^{- 31}$ kg and charge $1.6 \times 10^{- 19}$ C) is moving with speed $c / 100$ ( $c$ = 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 $c = 3 \times 10^{- 8}$ ${ms}^{- 1}$ ) [2025]

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

2 $\vec{E}$ is parallel to $\vec{B}$ and its magnitude is $27 \times 10^{4}$ $V$ $m^{- 1}$

3 $\vec{E}$ is perpendicular to $\vec{B}$ and its magnitude is $27 \times 10^{4}$ $V$ $m^{- 1}$

4 $\vec{E}$ is perpendicular to $\vec{B}$ and its magnitude is $27 \times 10^{2}$ $V$ $m^{- 1}$

Ans.
(4)

For an electron to move undeflected through perpendicular electric and magnetic fields, the net force on it must be zero.

i.e.,   $F_{B} = F_{E}$

or    $q v B = q E$

or $E = v B = \frac{c}{100} \times 9 \times 10^{- 4}$

   $= \frac{3 \times 10^{8}}{100} \times 9 \times 10^{- 4} = 27 \times 10^{2}$ $V/m$

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