Contents
Physics Topics can also be used to explain the behavior of complex systems, such as the stock market or the dynamics of traffic flow.
What is the Use of Cyclotron? What is Lorentz Force?
The cyclotron devised in 1932 by Professor E O Lawrence at the Berkely Institute, California is a powerful particle accelerator for accelerating positively charged particles such as protons, α particles, etc., to very high energies so that they can be used in disintegration experiments.
Description: It consists of two cylindrical shells of copper having semicircular cross section with same diameter and height. They are open towards their diameter and all other sides are closed. The diameters of the two shells are much larger than their heights [Fig.]. They are arranged side by side in such a way that a small gap exists in between their diameters. Each is called ‘dee’ on account of its shape like the letter D. Two pole pieces of a strong electromagnet are placed above and below of the dees in such a way that a uniform magnetic field acts perpendicular to the plane of the dees [A magnetic field \(\vec{B}\) directed upwards is shown in Fig.].
An alternating potential of the order of 105 V and of high frequency (about 106HZ) is applied between the dees. So the dees act as the two electrodes of the source of potential. An ion source S is located near the centre of the dees and it supplies the positive ions to be accelerated. At the periphery of the dees an auxiliary negative electrode deflects the accelerated ions on to the target to be bombarded. The whole space inside the dee is evacuated to a pressure of about 10-6 mm of mercury. If the whole arrangement is seen horizontally, it will look like the Fig.
Principle of action: Suppose that a positive ion (e.g., hydrogen ion H+ or helium ion He++) emerges from the ion sources S. Generally these ions are produced by bombarding gas molecules with high velocity electrons. The positive ion produced at S will be attracted into whichever dee happens to be negative at that moment. Due to this attractive force velocity of the ion increases and it enters the dee. Inside the dee there is no electric field. Owing to the magnetic field which is perpendic-ular to the plane of the dee the ion describes a semicircular path. The radius of the path is obtained from equation (1) of the sec-tion 1.6.1. The ion moves in this path with a constant speed. At the end of the semicircular path when the ion arrives at the gap between the dees, then the other dee should become negative. Then the ion is again attracted, its velocity increases and it enters the other dee. From equation
1. it is understood that due to increase of velocity v, the radius r of the semicircular path will also increase. But from the equation
2. of section 1.6.1 it is seen that in spite of increase of velocity and radius, the time taken to describe each of the semicircular paths remains the same. In this way the ion at last reaches the outer edge of the dee and comes out with a high velocity. To increase the velocity of the ion, the magnetic field (\(\vec{B}\)) may be increased (see equation (1) of section 1.6.1).
Resonance condition: It is obvious that the condition of proper operation of a cyclotron is that the time taken by the ion to traverse a semicircular path will be equal to half of the time period of the alternating potential. So frequency of the applied alternating potential difference n0 must be equal to the fre-quency of revolution of the ion, i.e., cyclotron frequency n. From equation (3) of section 1.6.1 we can write,
\(\frac{1}{2 \pi}\left(\frac{q}{m}\right) B\) = n0 …… (1)
This equation is called resonance condition of cyclotron. In practice the value of n0 is kept fixed and the magnetic field B is varied until the above condition is satisfied.
Kinetic energy of the accelerated particle: if v be the velocity of a charged particle of mass m and charge q and if it moves perpendicular to a magnetic field B, the radius of its circular path according to the equation (1) of section 1.6.1 is,
r = \(\frac{m v}{q B}\) or, v = \(\frac{q B r}{m}\)
If R be the radius of a dee of the cyclotron, then the velocity of the charged particle ejected from the outlet is given by,
vo = \(\frac{q B R}{m}\)
So, kinetic energy of, the particle,
E = \(\frac{1}{2} m v_0^2\) = \(\frac{q^2 B^2 R^2}{2 m}\) ….. (2)
Disadvantages:
i) If the velocity of the charged particle is high enough to be comparable to the velocity of light, the mass of the particle does not remain constant. According to the theory of rela-tivity m increases with the increase of velocity. So the res-onance condition according to equation (1) is violated and the cyclotron does not function. For an electron the relativ-istic increase of mass is much greater. So the electrons very quickly get out of step. Hence a cyclotron is not used for accelerating electrons.
ii) To make the charged particle sufficiently fast, dees of very large diameter are to be taken. This diameter may even exceed 100 m. The construction of an electromagnet of so large diameter is prohibitively expensive and technically complicated.
Remedies:
i) With the increase of mass due to relativity the frequency n0 of the alternating source may also be diminished in such a way that the product mn0 always remains constant. In that case the resonance condition is never violated. The machine with this arrangement is called synchro-cyclotron.
ii) Both B and n0 can be changed simultaneously in such a way that
1. the resonance condition is always satisfied and
2. in spite of increase of velocity of the charged particle, the radius r of its circular path remains unchanged. So a thin annular electromagnet of that radius will serve the purpose. Hence technical complications and expense may be reduced sufficiently. This machine is called synchrotron.
Path of a Charged Particle in a Uniform Electric Field
If a particle of charge +q experiences a force \(\vec{F}\) in a uniform electric field, from definition (see chapter : ‘Electric Field’), the electric field, \(\vec{E}\) = \(\frac{\vec{F}}{q}\) i.e., \(\vec{F}\) = \(q \vec{E}\) ….. (1)
If the mass of the particle is m, then acceleration
\(\vec{a}\) = \(\frac{\vec{F}}{m}\) = \(\frac{q \vec{E}}{m}\) …… (2)
Naturally, for negative charges, the acceleration (\(\vec{a}\)) will be opposite to \(\vec{E}\). Due to this acceleration the charged particle acquires different kinds of motion in different cases.
i) The charged particle is initially at rest : Initial velocity u = 0. Hence, after time t,
velocity of the particle, v = at = \(\frac{q E}{m} t\)
and displacement, s = \(\frac{1}{2} a t^2\) = \(\frac{q E}{2 m} t^2\)
ii) The charged particle enters with a velocity \(\overrightarrow{\boldsymbol{u}}\) along the electric field : Since the direction of acceleration is in the direction of \(\vec{E}\), after time t,
velocity of the particle, v = u + \(\frac{q E}{m} t\)
and the displacement, s = ut + \(\frac{1}{2} \frac{q E}{m} t^2\)
iii) The charged particle enters with a velocity \(\vec{u}\) perpendic-ular to the electric field: Let the uniform electric field act parallel to y-axis [Fig.] and the charged particle enters this electric field with a velocity \(\vec{u}\) at the point P along x -axis.
There is no component of \(\vec{E}\) along x -axis and hence the particle has no acceleration in this direction.
So, along x -axis, the particle possesses a uniform velocity u. If the distance covered by the particle in time t be x, then t = \(\frac{x}{u}\). Again, along y -axis, initial velocity of the particle = 0, but accel-eration, a = \(\frac{q E}{m}\). So, in time t, displacement of the particle along the y-axis is,
y = \(\frac{1}{2} a t^2\) = \(\frac{1}{2} \frac{q E}{m}\left(\frac{x}{u}\right)^2\) = \(\frac{q E}{2 m u^2} \cdot x^2\)
or, x2 = \(\frac{2 m u^2}{q E} y\) …. (3)
This is an equation of a parabola (in the form of x2 = 4ay)
So, the path of the particle will be parabolic.
Path of Charged Particle in Crossed Electric Field and Magnetic Field
Let a particle of charge +q and mass m enter an electromag-netic field at the point O along z-axis with a velocity \(\vec{v}\) [Fig.].
Electric field \(\vec{E}\) at that point O is along x-axis and magnetic field \(\vec{B}\) is along y -axis, i.e., \(\vec{v}\) \(\vec{E}\) and \(\vec{B}\) are mutually perpendicular.
Now, electric force acting on the particle along the x-axis = qE.
The, magnetic force acting on the particle = qvB; and the cross-product rule shows that this magnetic force acts along the negative direction of x -axis.
So, here the electric and the magnetic forces are oppositely directed. Keeping the directions of \(\vec{E}\) and \(\vec{B}\) unchanged, if these two forces are made equal in magnitude, the net force acting on the charged particle becomes zero. In this case, the charged particle continues its motion along z-axis with its initial velocity \(\vec{v}\) without suffering any deviation. The condition for this situation is, qE = qvB or, v = \(\frac{E}{B}\)
Suppose the electric field \(\vec{E}\) and magnetic field \(\vec{B}\) are acting simultaneously perpendicular to each other [Fig.]. Let a stream of charged particles enter the space in a direction perpendicular to both \(\vec{E}\) and \(\vec{B}\). If the speed of the charged particles are different, then only those particles whose speed is equal to the ratio – will pass through the hole S on the screen without any deflection. All other particles will be deflected from their path and will not reach the point S. This arrangement is known as velocity filter. In the figure only the green particle would pass through the hole S as its velocity is equal to \(\frac{E}{B}\).
This principle, known as the velocity selector, is utilised in the determination of
1. specific charge \(\left(\frac{q}{m}\right)\) of an electron by J J Thomson’s experiment and
2. mass of nucleus with the help of mass-spectrometer.
Lorentz Force
In an electromagnetic held, if a charge q enters with velocity \(\vec{v}\), the forces acting on it are
1. electric force, \(\vec{F}_e\) = q\(\vec{E}\) = [\(\vec{E}\) = electric field]
2. magnetic force, \(\vec{F}_m\) = \(q \vec{v} \times \vec{B}\) [\(\vec{B}\) = magnetic field]
These forces, \(\vec{F}_e\) and \(\vec{F}_m\) are called Loreniz electric force and Lorentz magnetic force, respectively. The resultant Lorentz force acting on the particle (charge),
\(\vec{F}\) = \(\vec{F}_e+\vec{F}_m\) = \(q \vec{E}+q \vec{v} \times \vec{B}\) = \(q(\vec{E}+\vec{v} \times \vec{B})\) ….. (1)
In CGS or Gaussian system:
\(\vec{F}_e\) = \(q \vec{E}\), \(\vec{F}_m\) = \(\frac{q}{c}(\vec{v} \times \vec{B})\)
So, the resultant Lorentz force, \(\vec{F}\) = \(q\left[\vec{E}+\frac{1}{c}(\vec{v} \times \vec{B})\right]\)
Remarks:
i) If the charged particle is at rest in an electromagnetic field, an electric force still acts on it, but as \(\vec{v}\) = 0 no magnetic force acts.
ii) For negatively charged particle (e.g., electron or negativeion) q is replaced by -q and hence the direction of each force (\(\vec{F}_e\), \(\vec{F}_m\) and \(\vec{F}\)) will be reversed.
Numerical Examples
Example 1.
A magnetic field of 0.40 T Is applied on a proton moving with a velocity of 5 × 106 m ᐧ s-1. The magnetic field acts at an angle 30: with the direction of velocity of the proton. What will be the acceleration of the proton? (mass of a proton = 1.6 × 10-27 kg)
Solution:
Force acting on the proton, F = qvBsinθ
∴ Acceleration of the proton \(=\frac{\text { force acting }}{\text { mass of proton }}\) = \(\frac{q v B \sin \theta}{m}\)
Here, q = 1.6 × 10-19C, v = 5 × 106m ᐧ s-1,
B = 0.40 T, θ = 30° and m = 1.6 × 10-27kg.
∴ Acceleration of the proton
= \(\frac{1.6 \times 10^{-19} \times 5 \times 10^6 \times 0.40 \times \sin 30^{\circ}}{1.6 \times 10^{-27}}\)
= 1014 m ᐧ s-2
Example 2.
An electron (mass = 9 × 10-31 kg, charge = -1.6 × 10-19 C) enters a magnetic field with velocity 106m ᐧ s-1 and starts rotating in a circular path of radius 10 cm. What is the value of the magnetic field?
Solution:
The electron is rotating in a circular path i.e. the direction of velocity of the electron is perpendicular to the direction of magnetic field. So the magnetic force = qvB which provides the necessary centripetal force for the circular motion.
∴ qvB = \(\frac{m v^2}{r}\)
[r = radius of the circular path = 10 cm = 0.1 m]
or, B = \(\frac{m v}{q r}\) = \(\frac{\left(9 \times 10^{-31}\right) \times 10^6}{\left(1.6 \times 10^{-19}\right) \times 0.1}\) = 5.6 × 10-5T
Example 3.
Two particles of equal charge are accelerated by applying the same potential difference and then allowed to enter a uniform magnetic field normally. If the particles keep revolving along circular paths of radii R1 and R2, determine the ratio of their masses.
Solution:
If the charge of each particle be q and the potential difference applied be V, kinetic energy acquired = qV
So, qV= \(\frac{1}{2} m_1 v_1^2\) = \(\frac{1}{2} m_2 v_2^2\)
or, \(\frac{v_1}{v_2}\) = \(\sqrt{\frac{m_2}{m_1}}\)
Again, magnetic force required to revolve along a circular path = centripetal force.
∴ For the first particle,
Example 4.
In a cyclotron the frequency of alternating current is 12 MHz and the radius of its dec is 0.53 m.
(i) What should be the operating magnetic field to accelerate protons?
(ii) What is the kinetic energy of the proton beam produced by the cyclotron? Given mass of proton = 1.67 × 10-27 kg and its charge = +1.6 × 10-19C.
Solution:
i) Condition of resonance,
\(\frac{1}{2 \pi}\left(\frac{q}{m}\right) B\) = n0 or, B = \(\frac{2 \pi n_0 m}{q}\)
Here, n0 = 12 MHz = 12 × 106 Hz = 12 × 106s-1
so, B = \(\frac{2 \times 3.14 \times\left(12 \times 10^6\right) \times\left(1.67 \times 10^{-27}\right)}{1.6 \times 10^{-19}}\)
= 0.79 Wb ᐧ m-2
ii) Kinetic energy of proton,
E = \(\frac{q^2 B^2 R^2}{2 m}\) = \(\frac{\left(1.6 \times 10^{-19}\right)^2 \times(0.79)^2 \times(0.53)^2}{2 \times\left(1.67 \times 10^{-27}\right)}\)
= 1.34 × 10-12J = \(\frac{1.34 \times 10^{-12}}{1.6 \times 10^{-19}} \mathrm{eV}\)
= 8.375 × 106 eV = 8.375 MeV
Example 5.
A beam of protons with velocity 4 × 105 m ᐧ s-1 enters a uniform magnetic field of 0.4 T at an angle of 60° to the magnetic field. Find the radius of the helical path of the proton beam and the time period of revolution. Also find the pitch of helix. Mass of proton = 1.67 × 10-27 kg.
Solution:
Mass of the proton, m = 1.67 × 10-27 kg; velocity of the proton, v = 4 × 105 m ᐧ s-1; charge, q = 1.6 × 10-19 C.
∴ Component of velocity perpendicular to field is vsinθ, the radius of the helical path,
r = \(\frac{m v \sin \theta}{q B}\) = \(\frac{\left(1.67 \times 10^{-27}\right)\left(4 \times 10^5\right) \sqrt{3}}{\left(1.6 \times 10^{-19}\right) \times 0.4 \times 2}\)
= 9 × 10-3 m = 0.9 cm
Time period of revolution,
T \(=\frac{2 \pi r}{v \sin \theta}\) = \(\frac{2 \times 3.14 \times 0.009 \times 2}{4 \times 10^5 \times \sqrt{3}}\) = 1.63 × 10-7S
Pitch of the helix,
p = vcosθ × T = 4 × 105 × \(\frac{1}{2}\) × 1.63 × 10-7
= 3.26 × 10-2m = 3.26 cm