Question 7 please The cyclotron (see figure) is a device used to accelerate elem

Question

Question 7 please

The cyclotron (see figure) is a device used to accelerate elementary particles such as protons to high speeds. Particles starting at point A with some initial velocity travel in circular orbits in the magnetic field B. The particles are accelerated to higher speeds each time they pass through the gap between the metal "dees, " where there is an electric field E. (There is no electric field inside the hollow metal dees.) The electric field changes direction each half cycle, owing to an AC voltage V = V_0 sin (2 pi f t), so that the particles are increased in speed at each passage through the gap. If the radius of the cyclotron is 5.610 times 10^-1 m, and the magnetic field strength 1.830 times 10^-1 T, what is the maximum kinetic energy of accelerated protons in eV? (The charge of a proton is 1.600 times 10^-19 C and the mass is 1.670 times 10^-27 kg. 1 electron volt is 1.600 times 10^-19 J) 5.04* 10^4 eV The figure shows a rectangular loop of wire of 120 turns, 14.0 by 34.0 cm. It carries a current of 2.50 A and is hinged at one side. What is the magnitude of the torque that acts on the loop, if it is mounted with its plane at an angle of 12.0 degree to the direction of B, which is uniform and equal to 0.200 T?

Explanation / Answer

maximum kinetic energy will correspond to speed at which radius of the semi circular path is equal to radius of the cyclotron.

let such speed be v m/s.

here speed is perpendicular to magnetic field.

hence magnitude of magnetic force=q*v*B

where q=charge of the proton

B=magnetic field strength

then equating magnetic force with centripetal force:

m*v^2/r=q*v*B

==>v=q*B*r/m

=1.6*10^(-19)*1.83*0.1*0.561/(1.67*10^(-27))

=9.836*10^6 m/s

then kinetic energy=0.5*mass*speed^2

=0.5*1.67*10^(-27)*(9.836*10^6)^2

=8.0784*10^(-14) J

=8.0784*10^(-14)/(1.6*10^(-19)) eV

=5.049*10^5 eV

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