A cyclotron is to accelerate protons to an energy of 5.4 MeV. The superconductin

Question

A cyclotron is to accelerate protons to an energy of 5.4 MeV. The superconducting electromagnet of the cyclotron produces a 3.5-T magnetic field perpendicular to the proton orbits.

Part A

When the protons have achieved a kinetic energy of 2.7 MeV , what is the radius of their circular orbit?

Express your answer using two significant figures.

6.8×102

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Correct

Part B

What is their angular speed?

Express your answer using two significant figures.

3.4×108

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Correct

Part C

When the protons have achieved a kinetic energy of 5.4MeV, what is the radius of their circular orbit?

Express your answer using two significant figures.

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Part D

What is their angular speed now?

Express your answer using two significant figures.

6.8×102

Explanation / Answer

rest mass of proton m=1.6748*10^-27 kg;
// charge of electron e= 1.602 E-19 C;
kinetic energy E=5.4MeV = 5.4e6*1.602e-19 J;
E =0.5m*v^2, hence v=(2E/m) =
= (2*5.4e6*1.602e-19 /1.6748e-27) 3.2e7 m/s;
non-relativistic proton,
centripetal force F=mw^2*r, where w=v/r is angular speed, r is radius of orbit;
force on the charge by magnetic field F=e*v*B =e*w*r*B;
F=F; mw^2*r = ewrB, hence w=eB/m =
= 1.602e-19 *3.5/1.6748e-27 =3.35e8 rad/s;

w does not depend on linear speed or energy!
v=(2E/m) = w*r
r= (2E/m)/w = E*(2*m)/(eB) =
= E * (2*1.6748e-27)/(1.602e-19 *3.5) =103220.5E;
r1 = 103220.5(2.7e6 *1.602e-19) =0.068 m;
r2=r1*2 =0.096 m;

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