Question

An electron is fired parallel to a uniform electric and magnetic fields acting simultaneously in the same direction. Then the electron

• A. Moves along a circular path
• B. Moves along a parabolic path
• C. Loses K.E
• D. Gains K.E

Answer 1

The correct option is C Loses K.E


From Lorentz's force equation,

$\overrightarrow{F}=q(\overrightarrow{E}+(\overrightarrow{v}\times \overrightarrow{B}\left)\right)$

$\because$ The electron $\left(e^{-}\right)$ is fired parallel to both $\overrightarrow{E}$ & $\overrightarrow{B}$

Magnetic force on the electron is,

$\overrightarrow{F_B}=q(\overrightarrow{v}\times \overrightarrow{B})=qvB\sin \theta [\because \theta =0]=q{v}_{0}B\sin 0^{\circ }=0$

Electric force on the electron is,

$\overrightarrow{F_E}=q\overrightarrow{E}=eE(-\hat{j})$

i.e $e^{-}$experiences force in opposite direction to the direction of $\overrightarrow{E}$ and gets accelerated in the same direction.

$a=\frac{-eE}{m}$

From kinematic equations of motion,

$v=u_0+atv=u_0-\frac{eE}{m}t$

i.e velocity of $e^{-}$ decreases with time $(v\downarrow )$

$\therefore KE=\frac{1}{2}m{v}^2\downarrow$ with time.

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option (c) is the correct answer.

Why this question ? Use of Lorentz's force in deciding velocity or kinetic energy of charged particle.