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Chapter 19: Problem 90

You want to build a cyclotron to accelerate protons to a speed of $3.0 \times 10^{7} \mathrm{m} / \mathrm{s} .$ The largest magnetic field strength you can attain is 1.5 T. What must be the minimum radius of the dees in your cyclotron? Show how your answer comes from Newton's second law.

Short Answer

Expert verified
Answer: The minimum radius of the dees in the cyclotron must be 2.09 meters.

Step by step solution

01

Identify the relevant equations and quantities

First, recall that the magnetic force acting on a moving charged particle is given by \(F_B = qvB\sin\theta\), where \(q\) is the charge, \(v\) is the velocity, \(B\) is the magnetic field strength, and \(\theta\) is the angle between the velocity and magnetic field. Since the protons move perpendicular to the magnetic field in a cyclotron, the angle is 90 degrees, and \(\sin\theta=1\). The magnetic force then becomes \(F_B = qvB\). Also, recall that the centripetal force acting on a particle moving in a circle of radius \(r\) is given by \(F_c = \frac{mv^2}{r}\), where \(m\) is the mass of the particle.
02

Apply Newton's second law

According to Newton's second law, the net force acting on an object is equal to its mass times its acceleration (\(F=ma\)). In this case, we will set the magnetic force equal to the centripetal force, as these are the only significant forces acting on the protons: \(F_B = F_c \Rightarrow qvB = \frac{mv^2}{r}\)
03

Solve for the minimum radius

We want to find the minimum radius, \(r\), of the dees, so we will rearrange the equation from Step 2 to solve for \(r\): \(r = \frac{mv}{qB}\) Now we can plug in the given values for the velocity and magnetic field strength, as well as the known values of the mass and charge of a proton (\(m_p = 1.67 \times 10^{-27} \mathrm{kg}\), \(q_p = 1.6 \times 10^{-19} \mathrm{C}\)): \(r_{min} = \frac{(1.67 \times 10^{-27}\, \mathrm{kg})(3.0 \times 10^{7}\, \mathrm{m} / \mathrm{s})}{(1.6 \times 10^{-19}\, \mathrm{C})(1.5\, \mathrm{T})}\)
04

Calculate the minimum radius

Lastly, perform the calculation: \(r_{min} = \frac{5.01 \times 10^{-20} \, \mathrm{kg\,m/s}}{2.4 \times 10^{-19}\, \mathrm{C}\,\mathrm{T}} = 2.09 \, \mathrm{m}\) So, the minimum radius of the dees in the cyclotron must be 2.09 meters.

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Most popular questions from this chapter

A long straight wire carries a current of 3.2 \(A\) in the positive \(x\)-direction. An electron, traveling at \(6.8 × 10 6\) m/s in the positive \(x\) -direction, is 4.6 cm from the wire. What force acts on the electron?
An early cyclotron at Cornell University was used from the 1930 s to the 1950 s to accelerate protons, which would then bombard various nuclei. The cyclotron used a large electromagnet with an iron yoke to produce a uniform magnetic field of \(1.3 \mathrm{T}\) over a region in the shape of a flat cylinder. Two hollow copper dees of inside radius \(16 \mathrm{cm}\) were located in a vacuum chamber in this region. (a) What is the frequency of oscillation necessary for the alternating voltage difference between the dees? (b) What is the kinetic energy of a proton by the time it reaches the outside of the dees? (c) What would be the equivalent voltage necessary to accelerate protons to this energy from rest in one step (say between parallel plates)? (d) If the potential difference between the dees has a magnitude of $10.0 \mathrm{kV}$ each time the protons cross the gap, what is the minimum number of revolutions each proton has to make in the cyclotron?
Two identical bar magnets lic onatable along a straight line with their north poles facing each other. Sketch the magnetic field lines.
An electron moves at speed \(8.0 \times 10^{5} \mathrm{m} / \mathrm{s}\) in a plane perpendicular to a cyclotron's magnetic field. The magnitude of the magnetic force on the electron is \(1.0 \times 10^{-13} \mathrm{N}\) What is the magnitude of the magnetic field?
A uniform magnetic ficld points north; its magnitude is 1.5 T. A proton with kinetic energy \(8.0 \times 10^{-13} \mathrm{J}\) is moving vertically downward in this field. What is the magnetic force acting on it?
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