Q177 / 180 · KCET 2010 · Physics
An electron enters the space between the plates of a charged capacitor as shown. The charge density on the plate is $\sigma$. Electric intensity in the space between the plates is E. A uniform magnetic field $B$ also exists in the space perpendicular to the direction of $E$
The electron moves perpendicular to both $\overrightarrow{\mathrm{E}}$ and $\overrightarrow{\mathbf{B}}$ without any change in direction. The time taken by the electron to travel a distance $l$ in the space is
- A$\frac{\sigma l}{\varepsilon_{0} \mathrm{~B}}$
- B$\frac{\sigma \mathrm{B}}{\varepsilon_{0} l}$
- Correct answerC$\frac{\varepsilon_{0} l \mathrm{~B}}{\sigma}$
- D$\frac{\varepsilon_{0} l}{\sigma \mathrm{B}}$
Solution
Here, magnetic force $=$ electrostatic force
$$
\begin{aligned}
q v B &=q E \\
v &=\frac{E}{B} \\
&=\frac{\sigma}{\varepsilon_{0} B}
\end{aligned}
$$
The time taken by electron to travel a distance 1 in that space
$$
\begin{aligned}
t=\frac{1}{v} &=\frac{1}{\frac{\sigma}{\varepsilon_{0} B}} \\
&=\frac{\varepsilon_{0} l B}{\sigma}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.