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 step by step, we will analyze the forces acting on the cathode rays in both scenarios.

### Step 1: Understand the forces acting on the cathode rays
When cathode rays move in a circular path, the centripetal force required to keep them in that path is provided by the electric force acting on them. The centripetal force \( F_c \) is given by:
\[ F_c = \frac{mv^2}{R} \]
where \( m \) is the mass of the cathode rays, \( v \) is their velocity, and \( R \) is the radius of the circular path.

### Step 2: Relate the centripetal force to the electric force
The electric force \( F_e \) acting on the charged particles in the electric field \( E \) is given by:
\[ F_e = qE \]
where \( q \) is the charge of the cathode rays.

Since the electric force provides the centripetal force, we can equate the two:
\[ \frac{mv^2}{R} = qE \]

### Step 3: Substitute the known values for the first scenario
From the problem, we know:
- Initial velocity \( v = 10^6 \, \text{m/s} \)
- Radius \( R = 1 \, \text{m} \)
- Electric field \( E = 300 \, \text{V/cm} = 30000 \, \text{V/m} \) (conversion from cm to m)

Substituting these values into the equation:
\[ \frac{m(10^6)^2}{1} = q(30000) \]
This simplifies to:
\[ m \cdot 10^{12} = 30000q \quad \text{(Equation 1)} \]

### Step 4: Analyze the second scenario with doubled velocity
Now, the problem states that the velocity is doubled, so the new velocity \( v' = 2v = 2 \times 10^6 \, \text{m/s} \).

The centripetal force equation for the new velocity becomes:
\[ \frac{m(2v)^2}{R} = qE' \]
Substituting \( v' \):
\[ \frac{m(2 \times 10^6)^2}{1} = qE' \]
This simplifies to:
\[ \frac{m \cdot 4 \cdot 10^{12}}{1} = qE' \quad \text{(Equation 2)} \]

### Step 5: Relate the two equations
From Equation 1, we have:
\[ m \cdot 10^{12} = 30000q \]
From Equation 2:
\[ m \cdot 4 \cdot 10^{12} = qE' \]
Now, we can express \( E' \) in terms of \( E \):
\[ E' = 4 \cdot (30000) = 120000 \, \text{V/m} \]
Converting back to \( \text{V/cm} \):
\[ E' = 1200 \, \text{V/cm} \]

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