Question

An electron moves along a straight line as shown inside a charged parallel plate capacitor. The space between the plates is filled with constant magnetic field of induction $\overrightarrow{B}$. Time taken for the straight line motion of the electron in the capacitor is

• A. $\frac{e\sigma }{{\epsilon }_{0}lB}$
  
• B. $\frac{{\epsilon }_{0}lB}{\sigma }$
  
• C. $\frac{e\sigma }{{\epsilon }_{0}B}$
  
• D. $\frac{{\epsilon }_{0}B}{e\sigma }$
  

Answer 1

The correct option is B $\frac{{\epsilon }_{0}lB}{\sigma }$


The net force acting on the electron must be zero as it moves with constant velocity along the straight line.
$\Rightarrow$ $\overrightarrow{F_{net}}=\overrightarrow{F_e}+\overrightarrow{F_m}=0\Rightarrow \left|\overrightarrow{F_e}\right|=\left|\overrightarrow{F_m}\right|\Rightarrow eE=evB\Rightarrow v=\frac{E}{B}=\frac{\sigma }{{\epsilon }_{0}B}[E=\frac{\sigma }{{\epsilon }_0}]$
$\therefore$ The time taken for the motion inside the capacitor $t=\frac{l}{v}=\frac{{\epsilon }_{0}lB}{\sigma }$ .