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Problem : A frequency of an electric oscillator is 10 MHz. What should be the magnitude of magnetic field for accelerating doubly ionized alpha particle? Assume mass of alpha particle 4 times that of proton.

Solution : The frequency of cyclotron is :

ν = q B 2 π m B = 2 π m ν q

Putting values,

B = 2 X 3.14 X 4 X 1.66 X 10 - 27 X 10 X 10 6 2 X 1.6 X 10 - 19 B = 1.3. T

Energy of charged particle

The energy of the finally accelerated particle corresponds to the speed when it travels in the outermost semicircular path having radius equal to that of Dees.

R = m v max q B v max = q B R m K max = 1 2 m v max 2 = q 2 B 2 R 2 2 m

Problem : Compare the final velocities of a proton particle and ionized deuteron when accelerated by a cyclotron. It is given that radius of cyclotron is 0.3 m and magnetic field is 2 T. Assume mass of deuteron twice that of the proton.

Solution : Let subscripts 1 and 2 correspond to proton and deuteron respectively. Note that deuteron is an isotope of hydrogen comprising of 1 proton and 1 neutron in the nucleus. The ionized deuteron thus carries one electronic positive charge same as proton. Now, final velocity of the charged particle accelerated by cyclotron is given as :

v max = q B R m

Hence,

v max1 v max2 = q 1 B 1 R 1 m 2 q 2 B 2 R 2 m 1

But R 1 = R 2 , q 1 = q 2 , B 1 = B 2 and m 2 = 2 m 1 . Thus,

v max1 v max2 = 2

Numbers of revolutions

The kinetic energy of the charged particle is increased every time it comes in the gap between the Dees. The energy is imparted to the charged particle in “lumps”. By design of the equipment of cyclotron, it is also evident that the amount of energy imparted to the particle is equal at every instance it crosses the gap between Dees.

Since particle is imparted energy twice in a revolution, the increase in energy corresponding to one revolution is :

Δ E = 2 q V

Let there be N completed revolutions. Then total energy,

E = N Δ E = 2 q N V

Equating this with the expression obtained earlier for energy, we have :

2 q N V = q 2 B 2 R 2 2 m N = q 2 B 2 R 2 4 m q V

Magnetic field and energy

From the expression of kinetic energy of the accelerated particle, it is clear that kinetic energy of the charged particle increases with the magnitude of magnetic field – even though magnetic field is incapable to bring about change in speed of the particle being always perpendicular to the motion. It is so because increasing magnetic field reduces the radius of curved motion inside Dees. Therefore, there are greater numbers of revolutions before reaching to the periphery. See that numbers of completed revolutions are directly proportional to the square of magnetic field.

N = q 2 B 2 R 2 4 m q V

Greater numbers of revolutions result in greater numbers of times electrons are subjected to electrical potential difference in the gap between Dees. The maximum kinetic energy of the particle is :

K max = 2 q N V

As such, energy of the emerging particle increases for a given construction of cyclotron when magnetic field increases.

Potential difference and energy

Again, it is clear from the expression of kinetic energy of the accelerated particle that the energy of emerging particle from the cyclotron is independent of potential applied in the gap. It appears to contradict the fact that it is the electric force which accelerates the charged particle in the gap. No doubt, the greater potential difference results in greater electric force on the charged particle. This, in turn, results in greater acceleration of the particle and hence velocity. But then, particle begins to rotate in greater semicircle. This results in lesser numbers of rotations possible within the fixed extent of Dees. In other words, the greater potential difference results in greater acceleration but lesser numbers of opportunities for acceleration. Now,

Questions & Answers

how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
Idrissa Reply
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Sherica
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Uday
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
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Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
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ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
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No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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Cied
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I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
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Porter
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Yasmin
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Cesar
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Uday
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what is system testing?
AMJAD
preparation of nanomaterial
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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AMJAD
what is system testing
AMJAD
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Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
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Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
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Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
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Prasenjit
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Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, Electricity and magnetism. OpenStax CNX. Oct 20, 2009 Download for free at http://cnx.org/content/col10909/1.13
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