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

Options

A. `(B^(2))/(2VE^(2))`
B. `(2VB^(2))/(E^(2))`
C. `(2VE^(2))/(B^(2))`
D. `(E^(2))/(2VB^(2))`

Answer

To solve the problem of finding the specific charge of cathode rays when subjected to crossed electric (E) and magnetic (B) fields, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Forces Acting on the Cathode Rays**:
- When the cathode rays are subjected to electric and magnetic fields, they experience forces due to both fields.
- The force due to the electric field (Fe) is given by:
\[ F_e = eE \]
where \( e \) is the charge of the electron and \( E \) is the electric field strength.
- The force due to the magnetic field (Fm) is given by:
\[ F_m = e(vB) \]
where \( v \) is the velocity of the cathode rays and \( B \) is the magnetic field strength.

2. **Set the Forces Equal**:
- Since the beam is not deflected, the forces must be equal:
\[ F_e = F_m \]
Therefore,
\[ eE = evB \]
- We can cancel \( e \) from both sides (assuming \( e \neq 0 \)):
\[ E = vB \]

3. **Solve for Velocity (v)**:
- Rearranging the equation gives us:
\[ v = \frac{E}{B} \]
- This is our first equation.

4. **Use the Law of Conservation of Energy**:
- The kinetic energy (KE) of the cathode rays is given by:
\[ KE = \frac{1}{2}mv^2 \]
- The potential energy (PE) due to the electric field is given by:
\[ PE = eV \]
- According to the conservation of energy:
\[ \frac{1}{2}mv^2 = eV \]

5. **Express e/m**:
- Rearranging the energy equation gives:
\[ mv^2 = 2eV \]
- Dividing both sides by \( m \) gives:
\[ v^2 = \frac{2eV}{m} \]
- Thus,
\[ \frac{e}{m} = \frac{v^2}{2V} \]
- This is our second equation.

6. **Substitute the Value of v**:
- From our first equation, substitute \( v = \frac{E}{B} \) into the equation for \( \frac{e}{m} \):
\[ \frac{e}{m} = \frac{\left(\frac{E}{B}\right)^2}{2V} \]
- Simplifying gives:
\[ \frac{e}{m} = \frac{E^2}{2VB^2} \]

7. **Final Expression for Specific Charge**:
- The specific charge \( \frac{e}{m} \) can also be expressed in terms of \( E \) and \( B \):
\[ \frac{e}{m} = \frac{E^2}{2VB^2} \]
- This matches with option 4:
\[ \frac{e^2}{2vB^2} \]

### Conclusion:
The specific charge of the cathode rays is given by:
\[ \frac{e}{m} = \frac{e^2}{2vB^2} \]
Thus, the correct option is option number 4: \( \frac{e^2}{2vB^2} \).