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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

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