Cathode rays of velocity `10^(6)ms^(-1)` d escribe an approximate circular path of the radius 1m in an electric field `"300 V cm"^(=-1)`. If the velocity of cathode rays are doubled. The value of electric field so that the rays describe the same circular path, will be

Options

A. `"120 V cm"^(-1)`
B. `"600 V cm"^(-1)`
C. `"1200 V cm"^(-1)`
D. `"12000 V cm"^(-1)`

Answer

To solve the problem, we will analyze the motion of cathode rays in an electric field and derive the necessary conditions for them to describe a circular path.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- We have cathode rays moving with an initial velocity \( V = 10^6 \, \text{m/s} \).
- The radius of the circular path is \( R = 1 \, \text{m} \).
- The electric field strength is \( E = 300 \, \text{V/cm} = 30000 \, \text{V/m} \) (converting to SI units).
- We need to find the new electric field \( E' \) when the velocity is doubled to \( 2V \).

2. **Centripetal Force and Electric Force:**
- The centripetal force required to keep the cathode rays in a circular path is given by:
\[ F_c = \frac{mv^2}{R} \]
- The electric force acting on the charged cathode rays in the electric field is given by:
\[ F_e = qE \]
- For the rays to move in a circular path, these two forces must be equal:
\[ \frac{mv^2}{R} = qE \]

3. **Setting Up the Equation for Initial Conditions:**
- For the initial conditions (velocity \( V \)):
\[ \frac{mV^2}{R} = qE \]
- Rearranging gives:
\[ E = \frac{mV^2}{qR} \]

4. **Setting Up the Equation for New Conditions:**
- When the velocity is doubled (\( 2V \)):
\[ \frac{m(2V)^2}{R} = qE' \]
- This simplifies to:
\[ \frac{4mV^2}{R} = qE' \]

5. **Relating the Two Electric Fields:**
- From the two equations, we can relate \( E' \) to \( E \):
\[ \frac{4mV^2}{R} = qE' \quad \text{and} \quad \frac{mV^2}{R} = qE \]
- Dividing the second equation from the first:
\[ \frac{4mV^2}{R} \div \frac{mV^2}{R} = \frac{qE'}{qE} \]
- This simplifies to:
\[ 4 = \frac{E'}{E} \]
- Thus, we find:
\[ E' = 4E \]

6. **Calculating the New Electric Field:**
- Substitute the initial electric field:
\[ E' = 4 \times 300 \, \text{V/cm} = 1200 \, \text{V/cm} \]

### Final Answer:
The new electric field \( E' \) required for the cathode rays to describe the same circular path when their velocity is doubled is \( 1200 \, \text{V/cm} \).

---