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A beam of cathode rays is subjected to crossed electric (E) and magnetic (B) fields. The fields are adjusted such that the beam is not deflected. The specific charge of the cathode rays is given by (where V is the potential difference between cathode and anode)
A
$\frac{E^2}{2VB^2}$
B
$\frac{B^2}{2VE^2}$
C
$\frac{2VB^2}{E^2}$
D
$\frac{2VE^2}{B^2}$
Detailed Solution
The beam is undeflected when the electric and magnetic forces balance: $eE = evB$, so $v = \frac{E}{B}$
The electrons gain their speed from the accelerating potential V: $eV = \frac{1}{2}mv^2$
$\frac{e}{m} = \frac{v^2}{2V}$
Substituting $v = \frac{E}{B}$: $\frac{e}{m} = \frac{E^2}{2VB^2}$
The electrons gain their speed from the accelerating potential V: $eV = \frac{1}{2}mv^2$
$\frac{e}{m} = \frac{v^2}{2V}$
Substituting $v = \frac{E}{B}$: $\frac{e}{m} = \frac{E^2}{2VB^2}$