Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A beam of cathode rays is subjected to crossed Electric (E) and magnetic fields (B). The fields are adjusted such that the beam is not deflected. The specific charge of the cathode rays is given by:

$\frac{B^2}{2VE^2}$

$\frac{2VB^2}{E^2}$

$\frac{2VE^2}{B^2}$

$\frac{E^2}{2VB^2}$ (where V is the potential difference between cathode and anode)

Step-by-Step Solution

Velocity Selector Condition: When a charged particle (electron) moves through crossed electric ( $E$ ) and magnetic ( $B$ ) fields without deflection, the electric force ( $F_e = eE$ ) is balanced by the magnetic force ( $F_m = evB$ ). $eE = evB \Rightarrow v = \frac{E}{B}$ .
Energy Conservation: The electrons are accelerated by a potential difference $V$ between the cathode and anode. The kinetic energy gained equals the potential energy lost: $\frac{1}{2}mv^2 = eV$ From this, the specific charge (charge-to-mass ratio) is: $\frac{e}{m} = \frac{v^2}{2V}$
Substitution: Substitute the expression for velocity ( $v = E/B$ ) into the specific charge equation: $\frac{e}{m} = \frac{(E/B)^2}{2V} = \frac{E^2}{2VB^2}$ .

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