CYCLOTRON WORKING PRINCIPLE

Cyclotron Working Principle

It is a device developed by Lawrence and Livingstone and is used to accelerate charged particles like protons and deuterons. This results in the production of high energy beam which is then used for artificial disintegration, etc.
It consists of two D shaped metal boxes (about 21″ in diameter) with their straight sides facing each other. There is a very small gap between these Dee’s faces across which a high potential difference (of order of 105 volts) is applied by a high frequency (10-15 mega cycles per second) oscillator. Thus an alternating electric field is established between Dee’s  during one half cycle, one dee is positive and the other is negative while during the next half cycle, the other is positive and one is negative. A source of particle (heavy hydrogen for producing deutrons) is placed between the faces. The whole apparatus is placed between the pole pieces NS of a very strong magnet.

Suppose a deuteron is at the centre of the gap and D1 is at negative potential compared to D2 then deuterons will be accelerated by the electric field between the dees towards D1. Since magnetic field B is perpendicular to the direction of motion of deuteron, a force will act on the particle given by
F = qv x B (where v and B are vectors) = qvB
Consequently, particle moves along a circular path whose radius is given by
So time period and frequency is independent of speed of charge and radius of circular path. If this frequency is equal to the frequency of the oscillator then by the time particle complete half revolution the polarity of the field will change. Deuteron will again be accelerated across the gap, its velocity will increase so that now it will move in a circular path of greater radius but its frequency will not be affected and the process of acceleration continues.
The particle is thus accelerated to gain high velocity. Its kinetic energy increases and when radius of the path reaches upto the value of the radius of the dee (R), beam is deflected by a negatively charged plate. Therefore with r=R, its maximum kinetic energy will be
So the necessary condition for accelerating the deuterons is
f=fo

where f0 is oscillator frequency. This is called resonance condition.

As velocity attained is high, we choose heavier particles e.g., protons, deuterons for acceleration in cyclotron. The reason is that at such high velocities, mass varies with velocity and f will then depend on velocity, upsetting the resonance condition. This variation in mass is more marked in the case of lighter particles. So cyclotron is suitable only for accelerating heavy particle like protons, deuterons a – particles etc.
Electrons can not be accelerated by cyclotrons because the mass of electron is very small and a small increase in energy of electron makes the electrons move with very high speed. As a result of it the electron go quickly out of step with oscillating electric field, it is for these reasons that other accelerating machines such as synchrotron, betatron (for accelerating electrons) have been developed. Cyclotron can not be used for accelerating uncharged particle like neutrons.

Theory and Working

The charged particle (say a positively charged proton) is released near mid point of the face of one of the Dees. Being in the electric field from one Dee to another, it is accelerated by the electric force in the direction of electric field. As the particle enters the adjoining Dee, the magnetic force, being perpendicular to it, renders the charged particle to move along a semicircular path within the Dee. By the time, it emerges again in the narrow gap separating the two Dees, the electrical polarity of Dees changes so that the particle is again accelerated again with an increase in speed.





But as the speed of the particle has increased, the radius of curvature of the semicircular path increases in accordance with the formula :



r = mv/Bq ( where B is magnetic field , m is mass , v is velocity , q is charge )


For given charge, mass and magnetic field, the radius is proportional to the speed. Clearly, the charged particle begins to move in a larger semicircular path after every passage through the gap. By the time particle reaches the gap successively, electric polarity of Dees keeps changing ensuring that the charged particle is accelerated with an increase in speed. This process continues till the charged particle reaches the periphery and exits through the guide with high energy and bombards a given target being investigated. The description of different segments of the path of accelerated particle is given here :


1: Path is a straight line. Particle is accelerated due to electric force. Speed and kinetic energy of the particle increase.


2: Path is a semicircular curve. Particle is accelerated due to magnetic force. This acceleration is centripetal acceleration without any change in speed and kinetic energy of the particle.


3: Path is a straight line. Particle is accelerated due to electric force in the direction opposite to the direction as in case 1. Speed and kinetic energy of the particle increase by same amount as in the case 1.


4: Path is a semicircular curve of greater radius of curvature due to increased speed. Particle is accelerated due to magnetic force. This acceleration is centripetal acceleration without any change in speed and kinetic energy of the particle.


5: Path is a straight line. Particle is accelerated due to electric force in the direction opposite to the direction as in case 1. Speed and kinetic energy of the particle increase by same amount as in the case 1 or 3.



We see that the particle follows consecutive larger semicircular path due to increase in the speed at the end of semicircular journey. The resulting path of charged particle, therefore, is a spiral path – not circular.

Cyclotron in India

Variable Energy Cyclotron Centre (VECC) is located in Calcutta, India.The Centre building itself houses a 224 cm cyclotron, was the first of its kind in India, having been operational since 1977-06-16. It provides proton, deuteron, alpha particle and heavy ion beams of various energies to other institutions..

Limitations of Cyclotron

Only when the speed of the circulating ion is less than 'c' the speed of light, we find the frequency of revolution to be independent of its speed.
At higher speeds, the mass of the ion will increase and this changes the time period of the ion revolution. This results in the ion lagging behind the electric field and it eventually loses by collisions against the walls of the dees.
*The cyclotron is suitable for accelerating heavy charged particles but not electrons.
*Cyclotrons cannot accelerate in charged particles.
*It is not suited for very high kinetic energy.

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Electric Field Due To A Line Charge

• Electric Field Due To A Line Charge

Consider a thin infinitely long straight line charge of linear charge density λ.
Let P be the point at a distance ‘a’ from the line. To find electric field at point P, draw a cylindrical surface of radius ‘a’ and length l.
If E is the magnitude of electric field at point P, then electric flux through the Gaussian surface is given by,

Φ = E × Area of the curved surface of a cylinder of radius r and length l

Because electric lines of force are parallel to end faces (circular caps) of the cylinder, there is no component of field along the normal to the end faces.

Φ = E × 2πal … (i)
According to Gauss theorem, we have

From equations (i) and (ii), we obtain

• Electric Field Due To An Infinite Plane Sheet Of Charge
  

Consider an infinite thin plane sheet of positive charge having a uniform surface charge density σon both sides of the sheet. Let P be the point at a distance ‘a’ from the sheet at which electric field is required. Draw a Gaussian cylinder of area of cross-section A through point P.
The electric flux crossing through the Gaussian surface is given by,

Φ = E × Area of the circular caps of the cylinder

Since electric lines of force are parallel to the curved surface of the cylinder, the flux due to electric field of the plane sheet of charge passes only through the two circular caps of the cylinder.

Φ = E × 2A … (i)
According to Gauss theorem, we have

Here, the charge enclosed by the Gaussian surface,
q = σA

From equations (i) and (ii), we obtain

• Electric Field Due To A Uniformly Charged Thin Spherical Shell
  

• When point P lies outside the spherical shell
  

Suppose that we have to calculate electric field at the point P at a distance r (r > R) from its centre. Draw the Gaussian surface through point P so as to enclose the charged spherical shell. The Gaussian surface is a spherical shell of radius r and centre O.
Let be the electric field at point P. Then, the electric flux through area element is given by,

Since is also along normal to the surface,
dΦ = E ds
∴ Total electric flux through the Gaussian surface is given by,

Now,

Since the charge enclosed by the Gaussian surface is q, according to Gauss theorem,

From equations (i) and (ii), we obtain

• When point P lies inside the spherical shell
  

In such a case, the Gaussian surface encloses no charge.
According to Gauss law,
E × 4πr² = 0
i.e., = E = 0 (r < R)

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BASIC PROPERTIES OF ELECTRIC CHARGES

Basic Properties of Electric Charges

• Additive nature of charges − The total electric charge on an object is equal to the algebraic sum of all the electric charges distributed on the different parts of the object. If q₁, q₂, q₃, … are electric charges present on different parts of an object, then total electric charge on the object, q = q₁ + q₂ + q₃ + …
  

• Charge is conserved − When an isolated system consists of many charged bodies within it, due to interaction among these bodies, charges may get redistributed. However, it is found that the total charge of the isolated system is always conserved.
  
• Quantization of charge − All observable charges are always some integral multiple of elementary charge, e (= ± 1.6 × 10⁻¹⁹ C). This is known as quantization of charge.
  

Coulomb’s Law

• Two point charges attract or repel each other with a force which is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.
  

Where, [In SI, when the two charges are located in vacuum]
− Absolute permittivity of free space = 8.854 × 10⁻¹² C² N⁻¹ m⁻²

We can write equation (i) as

• The force between two charges q₁ and q₂ located at a distance r in a medium may be expressed as
  

Where − Absolute permittivity of the medium

The ratio is denoted by ε_r, which is called relative permittivity of the medium with respect to vacuum. It is also denoted by k, called dielectric constant of the medium.

ε = kε₀

Coulomb’s Law in Vector Form

Consider two like charges q₁ and q₂ present at points A and B in vacuum at a distance r apart.
According to Coulomb’s law, the magnitude of force on charge q₁ due to q₂ (or on charge q₂ due to q₁) is given by,

Let
− Unit vector pointing from charge q₁ to q₂
− Unit vector pointing from charge q₂ to q₁
[is along the direction of unit vector ] …(ii)
[is along the direction of unit vector] …(iii)

∴Equation (ii) becomes

On comparing equation (iii) with equation (iv), we obtain

Forces between Multiple Charges
Principle of superposition − Force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to the other charges, taken one at a time. The individual forces are unaffected due to the presence of other charges.

Consider that n point charges q₁, q₂, q₃, … q_n are distributed in space in a discrete manner. The charges are interacting with each other. Let the charges q₂, q₃, … q_n exert forces on charge q₁. Then, according to principle of superposition, the total force on charge q₁ is given by,

If the distance between the charges q₁ and q₂ is denoted as r₁₂; and is unit vector from charge q₂ to q₁, then

Similarly, the force on charge q₁ due to other charges is given by,


Substituting these in equation (i),

Electric Field − It is the space around a charge, in which any other charge experiences an electrostatic force.
Electric Field Intensity − The electric field intensity at a point due to a source charge is defined as the force experienced per unit positive test charge placed at that point without disturbing the source charge.

Where,
→ Electric field intensity
Force experienced by the test charge q₀
Its SI unit is NC⁻¹.
Electric Field Due To a Point Charge

We have to find electric field at point P due to point charge +q placed at the origin such that
To find the same, place a vanishingly small positive test charge q₀ at point P.
According to Coulomb’s law, force on the test charge q₀ due to charge q is

If is the electric field at point P, then

The magnitude of the electric field at point P is given by,

Representation of Electric Field

Electric Field Due To a System of Charges

Consider that n point charges q₁, q₂, q₃, … q_n exert forces on the test charge placed at origin O.
Let be force due to i^th charge q_i on q₀. Then,

Where, r_i is the distance of the test charge q₀ from q_i
The electric field at the observation point P is given by,

If is the electric field at point P due to the system of charges, then by principal of superposition of electric fields,

Using equation (i), we obtain

Electric Field Lines
An electric line of force is the path along which a unit positive charge would move, if it is free to do so.

• Properties of Electric Lines of Force
  

• These start from the positive charge and end at the negative charge.
  
• They always originate or terminate at right angles to the surface of the charge.
  
• They can never intersect each other because it will mean that at that particular point, electric field has two directions. It is not possible.
  
• They do not pass through a conductor.
  
• They contract longitudinally.
  
• They exert a lateral pressure on each other.
  

• Representation of Electric Field Lines
  

• Field lines in case of isolated point charges
  

• Field lines in case of a system of two charges
  

  Continuous Charge Distribution

• Linear charge density − When charge is distributed along a line, the charge distribution is called linear.
  

Where,
λ → Linear charge density
q → Charge distributed along a line
L → Length of the rod

• Surface charge density
  

Where,
σ → Surface charge density
q → Charge distributed on area A

• Volume charge density
  The electric flux, through a surface, held inside an electric field represents the total number of electric lines of force crossing the surface in a direction normal to the surface.
  Electric flux is a scalar quantity and is denoted by Φ.
  
  
  SI unit − Nm² C⁻¹
  Gauss Theorem
  It states that the total electric flux through a closed surface enclosing a charge is equal to times the magnitude of the charge enclosed.
  
  However,
  ∴Gauss theorem may be expressed as
  
  Proof
  
  Consider that a point electric charge q is situated at the centre of a sphere of radius ‘a’.
  According to Coulomb’s law,
  
  Where, is unit vector along the line OP
  The electric flux through area element is given by,
  
  
  Therefore, electric flux through the closed surface of the sphere,
  
  It proves the Gauss theorem in electrostatics.
  

Where,
δ → Volume charge density
V → Volume of the conductor
q → Charge on conductor