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Chapter 19: Problem 26

The magnetic field in a cyclotron is \(0.50 \mathrm{T}\). What must be the minimum radius of the dees if the maximum proton speed desired is $1.0 \times 10^{7} \mathrm{m} / \mathrm{s} ?$

Short Answer

Expert verified
Answer: The minimum radius of the dees in the cyclotron must be approximately 0.104 m.

Step by step solution

01

Identify the relevant formula

The Lorentz force (F) acting on a charged particle of charge q moving with a velocity v in a magnetic field B is given by: F = q * (v x B) Where "x" denotes the cross product between velocity and magnetic field vectors. For a cyclotron, the charged particle moves in a circular path due to the magnetic field, so F also represents the centripetal force. Centripetal force is given by: F = (m * v虏) / R Where m is the mass of the particle, and R is the radius of the circular path.
02

Equate the two expressions for force

As the Lorentz force and the centripetal force both represent the force acting on the proton in a cyclotron, we can equate the two expressions: q * (v x B) = (m * v虏) / R
03

Simplify the equation

In a cyclotron, the magnetic field B is perpendicular to the velocity v. Therefore, the cross product simplifies to: v x B = v * B Now, we can substitute this back into the equation: q * (v * B) = (m * v虏) / R
04

Solve for the radius R

We can rearrange the equation to find the minimum radius of the dees (R): R = (m * v) / (q * B) For a proton, the mass (m) is 1.67 脳 10^(-27) kg, the charge (q) is 1.6 脳 10^(-19) C, the maximum proton speed (v) is 1.0 脳 10^(7) m/s, and the magnetic field (B) is 0.50 T. Using these values, we can now calculate R.
05

Calculate the minimum radius

Substitute the given values into the equation for R: R = (1.67 脳 10^(-27) kg * 1.0 脳 10^(7) m/s) / (1.6 脳 10^(-19) C * 0.50 T) R 鈮� 0.104 m The minimum radius of the dees in the cyclotron must be approximately 0.104 m.

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Most popular questions from this chapter

A square loop of wire of side \(3.0 \mathrm{cm}\) carries \(3.0 \mathrm{A}\) of current. A uniform magnetic field of magnitude \(0.67 \mathrm{T}\) makes an angle of \(37^{\circ}\) with the plane of the loop. (a) What is the magnitude of the torque on the loop? (b) What is the net magnetic force on the loop?
Prove that the time for one revolution of a charged particle moving perpendicular to a uniform magnetic field is independent of its speed. (This is the principle on which the cyclotron operates.) In doing so, write an expression that gives the period \(T\) (the time for one revolution) in terms of the mass of the particle, the charge of the particle, and the magnetic field strength.
Two identical bar magnets lic next to one another on a table. Sketch the magnetic ficld lines if the north poles are at the same end.
A long straight wire carries a current of 3.2 \(A\) in the positive \(x\)-direction. An electron, traveling at \(6.8 脳 10 6\) m/s in the positive \(x\) -direction, is 4.6 cm from the wire. What force acts on the electron?
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?
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