An electron enters the space between the plates of a charged capacitor as shown. The charge density on the plate is σ . 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 E and B without any change in direction. The time taken by the electron to travel a distance l in that space is

Question

An electron enters the space between the plates of a charged capacitor as shown. The charge density on the plate is σ . 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 E and B without any change in direction. The time taken by the electron to travel a distance l in that space is  (A)  (ϵ0 lB/σ) (B)  (ϵ l/σ B) (C)  (σ l /ϵ 0B) (D)  (σ B /ϵ 0 l)

Tardigrade Admin · Accepted Answer

Here, magnetic force = electrostatic force q v B =q E v =(E/B) =(σ/ε0 B) The time taken by electron to travel a distance l in that space t=(l/v) =(l/(σ)ε0 B) =(ε0 l B/σ)

$\frac{ϵ _{0} lB}{σ}$

$\frac{ϵ l}{σ B}$

$\frac{σ l}{ϵ _{0} B}$

$\frac{σ B}{ϵ _{0} l}$

Here, magnetic force $=$ electrostatic force
$q v B = qE$
$v = \frac{E}{B}$
$= \frac{σ}{ε _{0} B}$
The time taken by electron to travel a distance $l$ in that space
$t = \frac{l}{v} = \frac{l}{\frac{σ}{ε _{0} B}}$
$= \frac{ε _{0} lB}{σ}$

σϵ0​lB​

σBϵl​

ϵ0​Bσl​

ϵ0​lσB​

Here, magnetic force = electrostatic force qvB=qE v=BE​ =ε0​Bσ​ The time taken by electron to travel a distance l in that space t=vl​=ε0​Bσ​l​ =σε0​lB​

$\frac{ϵ _{0} lB}{σ}$

$\frac{ϵ l}{σ B}$

$\frac{σ l}{ϵ _{0} B}$

$\frac{σ B}{ϵ _{0} l}$

Here, magnetic force $=$ electrostatic force
$q v B = qE$
$v = \frac{E}{B}$
$= \frac{σ}{ε _{0} B}$
The time taken by electron to travel a distance $l$ in that space
$t = \frac{l}{v} = \frac{l}{\frac{σ}{ε _{0} B}}$
$= \frac{ε _{0} lB}{σ}$