An alternating electric field, of frequency v, is applied across the dees (radius =R) ofa cyclotron that is being used to accelerati protons (mass =m). The operating magnetic field (B) used in the cyclotron and the kinetic energy (K) of the proton beam, produced by it, are given by

Options

A. `B=(2pimv)/e` and `K=2mpi^(2)v^(2)R^(2)`
B. `B=(mv)/e` and `K=m^(2)pivR^(2)`
C. `B=(mv)/e` and `K=2mpi^(2)v^(2)R^(2)`
D. `B=(2pimv)/e` and `K=m^(2)piR^(2)`

Answer

To solve the problem, we need to determine the expressions for the operating magnetic field (B) and the kinetic energy (K) of protons in a cyclotron. Let's break it down step by step.

### Step 1: Understanding the Cyclotron Principle
A cyclotron accelerates charged particles, such as protons, using a magnetic field and an alternating electric field. The charged particle moves in a circular path due to the magnetic field, and the electric field accelerates it each time it crosses the gap between the dees.

### Step 2: Expression for the Magnetic Field (B)
The frequency (ν) of the alternating electric field is related to the motion of the charged particle in the magnetic field. The frequency of revolution (f) of a charged particle in a magnetic field is given by:

\[ f = \frac{qB}{2\pi m} \]

Where:
- \( q \) is the charge of the proton (approximately \( 1.6 \times 10^{-19} \) C),
- \( m \) is the mass of the proton,
- \( B \) is the magnetic field strength.

Given that the frequency of the electric field is \( \nu \), we can set \( f = \nu \):

\[ \nu = \frac{qB}{2\pi m} \]

Rearranging this equation to solve for the magnetic field \( B \):

\[ B = \frac{2\pi m \nu}{q} \]

### Step 3: Substituting the Charge of the Proton
Substituting the charge of the proton \( q = e \) (where \( e = 1.6 \times 10^{-19} \) C):

\[ B = \frac{2\pi m \nu}{e} \]

### Step 4: Expression for the Kinetic Energy (K)
The kinetic energy \( K \) of a charged particle in a cyclotron can be expressed using the velocity \( v \) of the particle. The velocity of a particle in a cyclotron is given by:

\[ v = \omega r \]

Where \( \omega \) is the angular frequency and \( r \) is the radius of the circular path. The angular frequency \( \omega \) is related to the frequency \( \nu \) by:

\[ \omega = 2\pi \nu \]

Thus, we can express the velocity as:

\[ v = 2\pi \nu r \]

The kinetic energy \( K \) is given by:

\[ K = \frac{1}{2} mv^2 \]

Substituting the expression for \( v \):

\[ K = \frac{1}{2} m (2\pi \nu r)^2 \]

Expanding this gives:

\[ K = \frac{1}{2} m (4\pi^2 \nu^2 r^2) = 2\pi^2 m \nu^2 r^2 \]

### Final Results
The expressions for the magnetic field and kinetic energy of the proton beam in the cyclotron are:

1. **Magnetic Field (B)**:
\[ B = \frac{2\pi m \nu}{e} \]

2. **Kinetic Energy (K)**:
\[ K = 2\pi^2 m \nu^2 r^2 \]