In a certain cyclotron a proton moves in a circle of radius 0.500 m. The magnitude of the magne

In a certain cyclotron a proton moves in a circle of radius...

In a certain cyclotron a proton moves in a circle of radius $0.500 \mathrm{~m}$. The magnitude of the magnetic field is $1.00 \mathrm{~T}$. (a) What is the oscillator frequency? (b) What is the kinetic energy of the proton, in electron-volts?

Key Concepts

Kinetic Energy and Energy Conversion

The kinetic energy of a moving particle is calculated using the formula K = ½mv². In problems involving charged particles in a cyclotron, once the speed is determined using the balance between the magnetic force and centripetal force, the kinetic energy can be computed. Often, this energy is converted from joules to electron-volts (using the conversion factor 1 eV = 1.602×10?¹? J) to align with common units used in particle physics.

Lorentz Force and Centripetal Force

The Lorentz force is the force experienced by a charged particle moving in a magnetic field and is given by F = qvB when the velocity is perpendicular to the magnetic field. This force provides the necessary centripetal force to keep the particle in a circular path, leading to the equation qvB = mv²/r. This relationship is fundamental in relating the speed of the particle to the radius of its circular path in a magnetic field.

Cyclotron Frequency

The cyclotron frequency (or oscillator frequency) is the frequency at which a charged particle orbits in a magnetic field. It is given by the formula f = (qB)/(2?m), where q is the charge of the particle, B is the magnetic field strength, and m is its mass. This frequency is independent of the radius and speed of the particle, assuming nonrelativistic conditions.

Cyclotron Motion

Cyclotron motion refers to the circular motion of a charged particle moving perpendicular to a uniform magnetic field. In this context, the magnetic Lorentz force acts as the centripetal force, leading to a circular path. The analysis of cyclotron motion is essential in understanding how charged particles are accelerated and steered in devices like cyclotrons.

Transcript

00:01 Hello everyone.

00:02 The question here is that a proton is moving inside a cyclotron in circles of radius 0 .5 meter under magnetic field of 1 tesla.

00:12 So in part a we have to find the oscillator frequency.

00:15 Part a, oscillator frequency equals to 1 by time period.

00:20 And the time period here, the formula of the time period would be 2 pi m by qb.

00:26 So substituting the formula here, qb by 2 pi mass of proton.

00:36 Q here is a charge, b here is the magnetic field.

00:40 Substituting the values, frequency of oscillator equals to 1 .6 charges 1 .6 into 10 to power minus 19 column.

00:52 The net unit here would be herds so we only to write the units in between into magnetic field would be 1 tesla, 2 into the 2 ,000, 3 .14 into mass of proton is 1 .67 into 10 to the power minus 27 kgis.

01:13 So the net answer would be approximately 1 .52 into 1030 power 7 hertz.

01:22 So the answer of the part a would be this and in part b we have to find the kinetic energy of the proton.

01:33 So in part b first we'll use the expression for the radius from there we'll derive the kinetic energy.

01:42 Mass of the proton radius equals to mass into velocity by charge into magnetic field.

01:49 Here we'll make use of an expression derived from kinetic energy equals to half mv square and mv equals to root of 2 m k.

02:05 We can clearly derive this expression from this.

02:09 So we'll substitute the numerator here as radius equals to under root of 2 into mass of proton into kinetic energy divided by q into b...

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