Question

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 (Where $V$ is the potential difference between cathode and anode)

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

Answer 1

The Correct Option is D

To solve this problem, we need to understand the principle of crossed electric and magnetic fields causing no deflection in a charged particle beam, such as cathode rays (electrons). This occurs when the electric force equals the magnetic force acting on the electrons, resulting in zero net force.

1. The condition for no deflection is when the electric force is equal to the magnetic force. Mathematically, this can be expressed as: qE = qvB where q is the charge of the electron, E is the electric field, v is the velocity of the electron, and B is the magnetic field.
2. From this condition, we can deduce the velocity of the electron: v = \frac{E}{B}
3. The specific charge of cathode rays is given by \frac{q}{m}, and we need to express this in terms of the given variables.
4. The relationship between the velocity, potential difference V, and specific charge can be derived as follows: The kinetic energy gained by the electron due to the potential difference V is given by: qV = \frac{1}{2}mv^2
5. Substitute v from step 2 into the kinetic energy equation: qV = \frac{1}{2}m\left(\frac{E}{B}\right)^2
6. Rearranging for the specific charge \frac{q}{m}: \frac{q}{m} = \frac{E^2}{2VB^2}
7. The correct option that matches this expression is: \frac {E^2}{2VB^2}.

Therefore, the specific charge of the cathode rays given the conditions is: \frac{E^2}{2VB^2}.