In a certain cyclotron a proton moves in a circle of radius 0.670 m. the magnitude of the magnetic field is 1.30 t. (a) what is the oscillator frequency? (b) what is the kinetic energy of the proton?

a) The formula for a cyclotron's frequency:

f = (qB)/(2πm)

f is the frequency, q is the particle's charge, B is the magnetic field strength, and m is the particle's mass.

Given values:

q = 1.6×10⁻¹⁹C (proton charge)

B = 1.30T

m = 1.67×10⁻²⁷kg (proton mass)

Plug in these values and solve for f:

f = (1.6×10⁻¹⁹×1.30)/(2π×1.67×10⁻²⁷)

f = 1.98×10⁷Hz

b) The kinetic energy of a particle in the cyclotron is:

KE = 0.5q²B²r²/m

q is the particle's charge, B is the magnetic field strength, r is the radius of the particle's path, and m is the particle's mass.

Given values:

q = 1.6×10⁻¹⁹C (proton charge)

B = 1.30T

r = 0.670m

m = 1.67×10⁻²⁷kg (proton mass)

Plug in the values and solve for KE:

KE = (0.5)(1.6×10⁻¹⁹)²(1.30)²(0.670)²/1.67×10⁻²⁷

KE = 5.81×10⁻¹²J

answer: a. same unit

explanation:

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y