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Chapter 19: Problem 25

The magnetic field in a cyclotron is \(0.360 \mathrm{T}\). The dees have radius \(82.0 \mathrm{cm} .\) What maximum speed can a proton achicve in this cyclotron?

Short Answer

Expert verified
Answer: The maximum speed a proton can achieve in this cyclotron is 5.71 x 10^7 m/s.

Step by step solution

01

Write down the given information and the cyclotron equation

We are given the magnetic field B = 0.360 T and the radius of the dees, r = 82.0 cm. The charge of a proton is q = +1.60 x 10^(-19) C, and its mass is m = 1.67 x 10^(-27) kg. The cyclotron equation relating these quantities is v = qBr/m.
02

Convert the radius of the dees to meters

Since the radius is given in centimeters, we need to convert it to meters by dividing by 100: r = 82.0 cm / 100 = 0.820 m.
03

Calculate the maximum speed using the cyclotron equation

We can now plug in the values for the charge of the proton (q), the magnetic field (B), the radius of the dees (r), and the mass of the proton (m) into the cyclotron equation v = qBr/m to find the maximum speed: v = (1.60 x 10^(-19) C)(0.360 T)(0.820 m) / (1.67 x 10^(-27) kg) v = 5.71 x 10^7 m/s The maximum speed a proton can achieve in this cyclotron is 5.71 x 10^7 m/s.

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Most popular questions from this chapter

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 straight wire segment of length \(0.60 \mathrm{m}\) carries a current of $18.0 \mathrm{A}$ and is immersed in a uniform external magnetic ficld of magnitude \(0.20 \mathrm{T}\). (a) What is the magnitude of the maximum possible magnetic force on the wire segment? (b) Explain why the given information enables you to calculate only the maximum possible force.
In a certain region of space, there is a uniform electric field \(\overrightarrow{\mathbf{E}}=3.0 \times 10^{4} \mathrm{V} / \mathrm{m}\) directed due east and a uniform magnetic field \(\mathbf{B}=0.080\) T also directed due east. What is the electromagnetic force on an electron moving due south at \(5.0 \times 10^{6} \mathrm{m} / \mathrm{s} ?\)
Electrons in a television's CRT are accelerated from rest by an electric field through a potential difference of \(2.5 \mathrm{kV} .\) In contrast to an oscilloscope, where the electron beam is deflected by an electric ficld, the beam is deflected by a magnetic field. (a) What is the specd of the electrons? (b) The beam is deflected by a perpendicular magnetic field of magnitude $0.80 \mathrm{T}$. What is the magnitude of the acceleration of the electrons while in the field? (c) What is the speed of the electrons after they travel 4.0 mm through the magnetic field? (d) What strength electric field would give the electrons the same magnitude acceleration as in (b)? (c) Why do we have to use an clectric ficld in the first place to get the electrons up to speed? Why not use the large acceleration due to a magnetic field for that purpose?
The conversion between atomic mass units and kilograms is $$1 \mathrm{u}=1.66 \times 10^{-27} \mathrm{kg}$$ Natural carbon consists of two different isotopes (excluding $^{14} \mathrm{C},$ which is present in only trace amounts). The isotopes have different masses, which is due to different numbers of neutrons in the nucleus; however, the number of protons is the same, and subsequently the chemical properties are the same. The most abundant isotope has an atomic mass of \(12.00 \mathrm{u} .\) When natural carbon is placed in a mass spectrometer, two lines are formed on the photographic plate. The lines show that the more abundant isotope moved in a circle of radius \(15.0 \mathrm{cm},\) while the rarer isotope moved in a circle of radius \(15.6 \mathrm{cm} .\) What is the atomic mass of the rarer isotope? (The ions have the same charge and are accelerated through the same potential difference before entering the magnetic field.)
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