An alternating electric field of frequency f is applied across the radius R of a cyclotron to accelerate protons ( mass m ). The operating magnetic field B used and K.E. of the proton beam produced by it are respectively (e = charge on proton)

Options

A. `(2pimf)/(e),2pi^(2)mf^(2)R^(2)`
B. `(2pi^(2)mf)/(e^(2)),4pi^(2)mf^(2)R^(2)`
C. `(pimf)/(e),pi^(2)mf^(2)R^(2)`
D. `(2pi^(2)m^(2)f^(2))/(e),2pi^(2)m^(2)f^(2)R^(2)`

Answer

To solve the problem, we need to determine the operating magnetic field \( B \) used in the cyclotron and the kinetic energy \( K.E. \) of the proton beam produced by it.

### Step 1: Understanding the Cyclotron
A cyclotron accelerates charged particles, such as protons, using a combination of an alternating electric field and a magnetic field. The protons move in circular paths due to the magnetic field, and their speed increases as they pass through the electric field.

### Step 2: Relationship between Frequency, Velocity, and Radius
The angular speed \( \omega \) of the proton in the cyclotron is given by:
\[ \omega = 2\pi f \]
where \( f \) is the frequency of the alternating electric field.

The linear speed \( v \) of the proton can be expressed in terms of the radius \( R \) and angular speed \( \omega \):
\[ v = R \omega = R (2\pi f) = 2\pi R f \]

### Step 3: Centripetal Force and Magnetic Force
In a magnetic field \( B \), the magnetic force acting on the proton provides the necessary centripetal force for circular motion. This relationship can be expressed as:
\[ \frac{mv^2}{R} = qvB \]
where \( m \) is the mass of the proton, \( q \) is the charge of the proton (denoted as \( e \)), and \( v \) is the speed of the proton.

### Step 4: Substituting for Velocity
Substituting \( v = 2\pi R f \) into the centripetal force equation gives:
\[ \frac{m(2\pi R f)^2}{R} = e(2\pi R f)B \]
Simplifying this, we have:
\[ m(2\pi R f)^2 = e(2\pi R f)B \cdot R \]
\[ m(2\pi f)^2 R = eB(2\pi f)R \]
Cancelling \( 2\pi f R \) from both sides (assuming \( R \neq 0 \)):
\[ m(2\pi f) = eB \]
Thus, we can solve for \( B \):
\[ B = \frac{2\pi mf}{e} \]

### Step 5: Kinetic Energy of the Proton
The kinetic energy \( K.E. \) of the proton can be calculated using the formula:
\[ K.E. = \frac{1}{2} mv^2 \]
Substituting \( v = 2\pi R f \):
\[ K.E. = \frac{1}{2} m (2\pi R f)^2 \]
\[ K.E. = \frac{1}{2} m (4\pi^2 R^2 f^2) \]
\[ K.E. = 2\pi^2 m R^2 f^2 \]

### Final Answers
1. The operating magnetic field \( B \) is:
\[ B = \frac{2\pi mf}{e} \]
2. The kinetic energy \( K.E. \) of the proton beam is:
\[ K.E. = 2\pi^2 m R^2 f^2 \]