An electron having mass m and kinetic energy E enter in uniform magnetic field B perpendicularly, then its frequency will be

Question

An electron having mass m and kinetic energy E enter in uniform magnetic field B perpendicularly, then its frequency will be (A) ( eE/ qvB) (B) ( 2 π m/ e B) (C) ( eB/2 π m ) (D) ( 2 m / e B E )

Tardigrade Admin · Accepted Answer

The frequency of revolution of charged particle in a perpendicular magnetic field is upsilon = (1/T) = ( 1/ 2 π r /v ) = ( v/ 2π r) = (v/2 π ) × ( e B /mv ) = (eB/2 π m ).

Q. An electron having mass m and kinetic energy $E$ enter in uniform magnetic field $B$ perpendicularly, then its frequency will be

$\frac{e E}{q v B}$

$\frac{2 πm}{e B}$

$\frac{e B}{2 πm}$

$\frac{2 m}{e BE}$

The frequency of revolution of charged particle in a perpendicular magnetic field is
$υ = \frac{1}{T} = \frac{1}{2 π r / v}$
$= \frac{v}{2 π r} = \frac{v}{2 π} \times \frac{e B}{m v}$
$= \frac{e B}{2 πm}$ .

Q. An electron having mass m and kinetic energy E enter in uniform magnetic field B perpendicularly, then its frequency will be

qvBeE​

eB2πm​

2πmeB​

eBE2m​