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

Part a)

`f = 1.46 * 10^7 Hz`

Part b)

`KE = 3.87 * 10^(-12) J`

Step-by-step explanation:

Part a)

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

`R = (mv)/(qB)`

now we have

`v = (qBR)/(m)`

now the frequency of oscillator is given as

`f = (v)/(2\pi R)`

`f = (qB)/(2\pi m)`

`f = ((1.6 * 10^(-19))(0.960))/(2\pi(1.67* 10^(-27)))`

`f = 1.46 * 10^7 Hz`

PART b)

now for kinetic energy of proton we will have

`KE = (1)/(2)mv^2`

`KE = (1)/(2)m((qBR)/(m))^2`

`KE = (q^2B^2R^2)/(2m)`

`KE = ((1.6 * 10^(-19))^2(0.960)^2(0.740)^2)/(2(1.67* 10^(-27)))`

`KE = 3.87 * 10^(-12) J`