- Home
- Standard 12
- 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 that space perpendicular to the direction of $E$. The electron moves perpendicular to both $\vec E$ and $\vec B$ without any change in direction. The time taken by the electron to travel a distance $\ell $ is the space is
$\frac{{\sigma \ell }}{{{\varepsilon _0}B}}$
$\frac{{\sigma B }}{{{\varepsilon _0 \ell }B}}$
$\frac{{{\varepsilon _0}\ell B}}{\sigma }$
$\frac{{{\varepsilon _0}\ell }}{{\sigma B}}$
Solution
Force on the electron due to the electric field $\mathrm{E}$ is
$\mathrm{F}_{\mathrm{E}}=(-\mathrm{e}) \mathrm{E}$
Force on the electron due to the magnetic field $B$ is
$\mathrm{F}_{\mathrm{B}}=(-\mathrm{e}) \mathrm{v} \mathrm{B}$
The electron will move in the fields undeflected, if these two forces are equal and opposite.
$\mathrm{eE}=\mathrm{evB} \quad$ or $\quad \mathrm{v}=\frac{\mathrm{E}}{\mathrm{B}}$
Electric field between the plates is $\mathrm{E}=\frac{\sigma}{\varepsilon_{0}}$
$\therefore \mathrm{v}=\frac{\sigma}{\varepsilon_{0} \mathrm{B}}$
The time taken by the electron to travel a distance $\ell$ in the space is $t=\frac{\ell}{V}=\frac{\ell}{\frac{\sigma}{\varepsilon_{0} B}}=\frac{\ell \varepsilon_{0} B}{\sigma}$
