Question

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

• A. $\frac{B^2}{2v{E}^2}$
• B. $\frac{2v{B}^2}{E^2}$
• C. $\frac{2v{E}^2}{B^2}$
• D. $\frac{E^2}{2v{B}^2}$

Answer 1

The correct option is D

$\frac{E^2}{2v{B}^2}$

Step 1: Given Data

Electric field $=E$

Magnetic field $=B$

Let a single charge be $e$.

Let the velocity with which the charge moves be $v$.

Step 2: Calculate the Velocity

We know that the total force due to the electric field is given as

$F_E=eE$

We know that the total force due to the magnetic field is given as

$F_B=evB$

Since the beam is not deflected the force due to the electric field and the magnetic field is equal.

$\therefore F_E=F_B$

Upon substituting the value we get,

$eE=evB$

$\Rightarrow v=\frac{E}{B}$

Step 3: Calculate the Specific Charge

Since the charge is moving with kinetic energy, the energy of the charge can be given as

$\frac{1}{2}m{v}^2=ev$

$\Rightarrow v=\sqrt{\frac{2ev}{m}}$

Upon substituting the value of $v$ we get,

$\frac{E}{B}=\sqrt{\frac{2ev}{m}}$

$\Rightarrow \frac{e}{m}=\frac{E^2}{2v{B}^2}$

We know that specific charge $=\frac{e}{m}$

Therefore, specific charge $=\frac{E^2}{2v{B}^2}$

Therefore, the correct answer is option (D).