Question

A beam of cathode rays is subjected to a crossed electric field$\left( E \right)$and magnetic field$\left( B \right)$. The fields are adjusted such that the beam is not deflected. The specific charge of cathode rays is given by:
$\text{A}\text{. }\dfrac{{{B}^{2}}}{2v{{E}^{2}}}$
$\text{B}\text{. }\dfrac{2v{{B}^{2}}}{{{E}^{2}}}$
$\text{C}\text{. }\dfrac{2v{{E}^{2}}}{{{B}^{2}}}$
$\text{D}\text{. }\dfrac{{{E}^{2}}}{2v{{B}^{2}}}$

Answer 1

Hint:
As the cathode ray is passing undeflected from the crossed electric and magnetic field, it means that the force exerted on the particles of ray by electric field is balanced by the force exerted on the particles of ray by magnetic field. Equating the two forces will give the value of specific charge of cathode rays.
Formula used:
${{F}_{e}}=Ee$
${{F}_{m}}=Bev$

Complete step-by-step answer:
Cathode rays are a beam of electrons which are emitted from the cathode of a high volume tube. The cathode is a negatively charged conductor, and the anode is a positively charged conductor. Electrons, which carry negative charge, flow off the cathode and are attracted towards the anode. A small hole in anode allows some electrons to pass through it, hence creating a beam of electrons known as Cathode rays.
Specific charge of an ion or subatomic particles is the ratio of the charge carried by the particle to its mass.
Cathode ray tubes use a focused beam of electrons deflected by the presence of crossed electric and magnetic fields to render an image on the screen.
In case the electron is not deflected, then the force due to electric field is balancing the force due to magnetic field,
${{F}_{m}}={{F}_{e}}$
Force exerted on the particles of cathode ray by the presence of electric field,
${{F}_{e}}=Ee$
Where,
$E$is the electric field intensity
$e$is the charge on electron
Force exerted on the particles of cathode ray by the presence of magnetic field,
${{F}_{m}}=Bev$
Where,
$B$is the magnetic field intensity
$e$is the charge on electron
$v$is the velocity of particles of cathode ray
As force due to electric field is equal to force due to magnetic field,
${{F}_{m}}={{F}_{e}}$
We get,
$Bev=Ee$
$v=\dfrac{E}{B}$
According to the law of conservation of energy, we have,
$\dfrac{1}{2}m{{v}^{2}}=eV$
$v=\sqrt{\dfrac{2eV}{m}}$
$\sqrt{\dfrac{2eV}{m}}=\dfrac{E}{B}$
$\dfrac{e}{m}=\dfrac{{{E}^{2}}}{2V{{B}^{2}}}$
Therefore, the specific charge of cathode rays is given by, $\dfrac{e}{m}=\dfrac{{{E}^{2}}}{2V{{B}^{2}}}$
Hence, the correct option is D.

Note:
Crossed electric and magnetic fields means that the electric field and magnetic fields are perpendicular to each other. Because of this only, force on cathode ray is balanced and it goes undeflected through the region. If the direction of electric and magnetic fields were in the same direction, the force on the particles of the ray would not have balanced and the ray would get deflected in the region.