Question

In a certain cyclotron a proton moves in a circle of radius 0.740 m. The magnitude of the magnetic field is 0.960 T. (a) What is the oscillator frequency? (b) What is the kinetic energy of the proton?

Answer 1

Answer:

Part a)

$f = 1.46 \times 10^7 Hz$

Part b)

$KE = 3.87 \times 10^{-12} J$

Explanation:

Part a)

As we know that radius of circular path of a charge moving in constant magnetic field is given as

$R = \frac{mv}{qB}$

now we have

$v = \frac{qBR}{m}$

now the frequency of oscillator is given as

$f = \frac{v}{2\pi R}$

$f = \frac{qB}{2\pi m}$

$f = \frac{(1.6 \times 10^{-19})(0.960)}{2\pi(1.67\times 10^{-27})}$

$f = 1.46 \times 10^7 Hz$

PART b)

now for kinetic energy of proton we will have

$KE = \frac{1}{2}mv^2$

$KE = \frac{1}{2}m(\frac{qBR}{m})^2$

$KE = \frac{q^2B^2R^2}{2m}$

$KE = \frac{(1.6 \times 10^{-19})^2(0.960)^2(0.740)^2}{2(1.67\times 10^{-27})}$

$KE = 3.87 \times 10^{-12} J$