A cyclotron with dee radius 53.0 cm is operated at an...

A cyclotron with dee radius 53.0 $\mathrm{cm}$ is operated at an oscillator frequency of 12.0 $\mathrm{MHz}$ to accelerate protons. (a) What magnitude $B$ of magnetic field is required to achieve resonance? (b) At that field magnitude, what is the kinetic energy of a proton emerging from the cyclotron? Suppose, instead, that $B=1.57 \mathrm{T}$ . (c) What oscillator frequency is required to achieve resonance now? (d) At that frequency, what is the kinetic energy of an emerging proton?

Transcript

00:01 Hello, the question is taken from physics and the question is a cyclotron with d radius is equal to 53 centimetre.

00:10 So given that r is equal to 53 centimetre and the value of a frequency is given as 12 megahertz.

00:20 So 12 megahertz is the frequency value and what magnitude of magnetic field is required to achieve resonance.

00:27 So what will be the value of b in order to achieve the? resonance and so let me evaluate first this value of b we know that in case of a cyclotron the magnetic force is balanced by centripetal force of acceleration so we will be cancelled out so from here r is equal to m b over e b okay or we can write b is equal to m b over e into r v is equal to r v is equal to to so we for that b is equal to mv over e into r and we we know that is equal to r into omega omega is the angular frequency so we is if v is equal to r omega in that case we can write b is equal to m omega over e i think i think that is fine when we will be can so m v over e b and b and b will be mv over e into r v is equal to r into omega am omega by a so that will be two pi m new over e so from here the value of magnetic field will be equal to two into 3 .14 and the mass of electron is equal to 9 .1 into 10 to the power minus 31 frequency is given to us that is 12 megahertz 12 into 10 to the power 6 divided by 1 .6 into 10 to the power minus 19 so from here the value of magnetic field will be equal to 6 .28 into 9 .1e minus 31 into 12 6 divided by 1 .6e minus 19 so the value of magnetic field will be equal to 4 .28 into 10 to for tesla and at what filmed magnitude the kinetic energy of proton emerging from the at that field magnitude what is the kinetic energy of proton that is a case of proton so that is not for electron okay so let me and take the mass of proton so if we take the mass of proton here we get it's a 1 .67 into 10 to the power minus 27 now let us again calculate its value 6 .28 into 1 .67e minus 27 into 12v6 divided by 1 .6 e minus 19 so that value will be equal to sorry and the value is 0 .78 675 tesla that is the required value of magnetic field and kinetic energy is equal to half m v square that is equal to half m the value of v from this equation is erb by m so that will be e square r square b square over m square one m will be cancelled out so that will be e square r square b square over 2m okay so kinetic energy from here will be equal to we know the value of er b and m so let us evaluate that value 1 .6e minus 19 square into the value of radius is 53 so multiply 0 .53 square into magnetic field value is 0 .78 675 square divided by 2 divided by mass of proton that is 1 .67 e minus 27 so that value will be 1 .6 .7.

05:04 So that value will be 1 1 .333 into 10 to the power minus 12 jule.

05:12 Let us convert into electron volt divided by 1 .60 minus 19...

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