Question

An electron is released from the origin at a place where a uniform electric field $\overrightarrow{E}$ and uniform magnetic field $\overrightarrow{B}$ exist along the negative y - axis and the negative z - axis respectively

• A. At time t the y-component of velocity of the electron becomes $u_y=\frac{E}{B}sin\omega t$ where $\omega =\frac{eB}{m}$.
• B. At $\frac{\pi m}{eB}$ the electron will have only x-component of velocity.
• C. At $t=\frac{2\pi m}{3em}$, the y-component of velocity becomes zero.
• D. The displacement along y-axis is $\frac{2Em}{e{B}^2}$ when the velocity of electron becomes perpendicular to the y-axis

Answer 1

Answer: (A), (B), (D)

The correct options are
A At time t the y-component of velocity of the electron becomes $u_y=\frac{E}{B}sin\omega t$ where $\omega =\frac{eB}{m}$.
B At $\frac{\pi m}{eB}$ the electron will have only x-component of velocity.
D The displacement along y-axis is $\frac{2Em}{e{B}^2}$ when the velocity of electron becomes perpendicular to the y-axis

Since $u_y=\frac{E}{B}sin\omega t$, where $\omega =\frac{eB}{m}$
(using Lorentz's force equation and solving it)
At $t=\frac{\pi m}{eB},u_y=0$, hence it will have only x component of velocity.
$y={\int }_0^{\frac{xm}{eB}}\frac{E}{B}sin\left(\omega t\right)dt=\frac{2Em}{e{B}^2}$