The figure below shows a schematic drawing of a cyclotron, which is used as part

Question

The figure below shows a schematic drawing of a cyclotron, which is used as particle accelerator, when particles of high speeds are required. A charged particle starts out at the central point. An alternating voltage is applied across the gap between the Dees. The particle crosses the gap and is accelerated by the potential difference across the gap. The particle then enters a uniform magnetic field which forces it on a circular path. After a half-circle the particle reaches the gap, and is again accelerated by the voltage which now points in opposite direction, i.e. again in direction of particle motion. This process is repeated and the particle is accelerated in each step.

A voltage V is applied across the gap. What is the speed of a particle with mass m and charge q, starting from rest, after it has crossed the gap for the first time?

What is the radius of curvature, when it enters the magnetic field B? How long does it take until the particle has completed the semicircle and reaches the gap the second time?

The particle is crossing the gap for the second time. What is its speed, when it enters the magnetic field after this second crossing? What are the radius of curvature and the time for completing the second semicircle?

The voltage must alternate each time the particle reaches the gap. Look at the results from questions b and c. What is the time between two alternating steps? Does it change with the increasing speed of the particle?

A cyclotron has a maximum radius R, at which the particle leaves the accelerator. What is the speed of the particle, when its path has a radius R? What is its energy? How many times has the particle crossed the gap?

A cyclotron is used to accelerate protons. The electric field between the gap is 50 kV. The magnetic field is 1T and the maximum radius is 50 cm. What is the speed of the proton when it leaves the cyclotron? How many times has it crossed the gap?

The figure below shows a schematic drawing of a cyclotron, which is used as particle accelerator, when particles of high speeds are required. A charged particle starts out at the central point. An alternating voltage is applied across the gap between the ½Dees½. The particle crosses the gap and is accelerated by the potential difference across the gap. The particle then enters a uniform magnetic field which forces it on a circular path. After a half-circle the particle reaches the gap, and is again accelerated by the voltage which now points in opposite direction, i.e. again in direction of particle motion. This process is repeated and the particle is accelerated in each step. A voltage V is applied across the gap. What is the speed of a particle with mass m and charge q, starting from rest, after it has crossed the gap for the first time? What is the radius of curvature, when it enters the magnetic field B? How long does it take until the particle has completed the semicircle and reaches the gap the second time? The particle is crossing the gap for the second time. What is its speed, when it enters the magnetic field after this second crossing? What are the radius of curvature and the time for completing the second semicircle? The voltage must alternate each time the particle reaches the gap. Look at the results from questions b and c. What is the time between two alternating steps? Does it change with the increasing speed of the particle? A cyclotron has a maximum radius R, at which the particle leaves the accelerator. What is the speed of the particle, when its path has a radius R? What is its energy? How many times has the particle crossed the gap? A cyclotron is used to accelerate protons. The electric field between the gap is 50 kV. The magnetic field is 1T and the maximum radius is 50 cm. What is the speed of the proton when it leaves the cyclotron? How many times has it crossed the gap?

Explanation / Answer

a)

The work done by the electric field, W = V*q

Now by conservation of energy, this Work done must be converted to Kinetic Energy

So, Vq = 0.5*mv^2

So, v = sqrt(2Vq/m)

b)

For circular motion,

the magnetic force must provide the necessary centripetal force

So, qvB = mv^2/r

So r = mv/qB ,<---------- , radius of curvature

c)

velocity = distance traveled in semicircle/time

So, time = distance/velocity = pi*r/v = pi*(mv/qB)/v = pi*m/qB <------ time for reaching the gap the second time

d)

Again, change in Kinetic energy = Vq

So, 0.5*m*(v'^2-v^2) = Vq

So, v' = sqrt(2Vq/m + v^2) <---------speed at the gap in the second time

e)

Now, r = mv'/qB = m*(sqrt(2Vq/m+v^2))/(qB) <-------radius

f)

time taken during 2nd crossing. t = pi*m/qB

As time does not depend on the velocity, so both the times in above will be same

So, no , it does not change with increasing speeds

g)

speed of the particle , when it has radius can be found out using the same relation as in part (b)

r= mv/qB

So, v = qBr/m

So, speed when radius = R, v = qBR/m <----------answer

Its energy, U = 0.5*mv^2 = 0.5*m*(qBR/m)^2 = 0.5*q^2B^2R^2/m <--------answer

Every time it crosses the gap its energy inceases by W = V*q amound

So, number of times it has crossed the gap = (0.5*(qBR)^2/m)/Vq

h)

V = 50*10^3 V

B = 1 T

R = 0.5 m

So, v = qBR/m

where q = charge of proton = 1.6*10^-19 C

m = 1.67*10^-27 kg

So, v = 1.6*10^-19*(1)*(0.5)/(1.67*10^-27) = 4.8*10^7 m/s <------answer (speed when it leaves the cyclotron)

Number of times it has crossed, = (0.5*(qBR)^2/m)/Vq

= (0.5*(1.6*10^-19*1*0.5)^2/(1.67*10^-27))/(50*10^3*1.6*10^-19)

= 239.5

= 239 times it has crossed

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