The maximum velocity to which a proton can be accelerated in a cyclotron of 10 MHZ frequency and radius 50 cm is

Question

The maximum velocity to which a proton can be accelerated in a cyclotron of 10 MHZ frequency and radius 50 cm is

The maximum velocity to which a proton can be accelerated in a cyclotron of 10 MHZ frequency and radius 50 cm is
A. `6.28xx10^(8) ms^(-1)`
B. `3.14xx10^(8) ms^(-1)`
C. `6.28xx10^(7) ms^(-1)`
D. `3.14 xx10^(7) ms^(-1)`

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

Related Questions

A 1.5m diameter cyclotron is used to accelerate proton to an energy of 8 MeV. The required magnetic field strength for this cyclotron is : - `(m_(p)=1

2 Answers 4 Views

The operating magnetic field for accelerating protons in a cyclotron oscillator having frequency of 12 MHz is `(q= 1.6 xx 10^(-19) C, m_(p) = 1.67 xx

2 Answers 4 Views

A proton is accelerating on a cyclotron having oscillating frequency of 11 MHz in external magnetic field of 1 T. If the radius of its dees is 55 cm,

2 Answers 4 Views

A radio can tune to any station to 7.3 MHz to 12 MHz band. The corresponding wavelength band is

2 Answers 4 Views

On a particular day, the maximum frequency reflected from the ionosphere is 10 MHz. On another day, it was found to increase to 11 MHz. Calculate the

2 Answers 4 Views

If the radius of the dees of cyclotron is r then the kinetic energy of a proton of mass m accelerated by the cycloton at an oscillating frequency v is

2 Answers 4 Views

a proton enters a perpendicular magnetic field of intensity b with speed v the proton moves along a circular path of radius r if a deuteron enter into the same magnetic field with same speed the radius of the circular trajectory will be

2 Answers 5 Views

The ratio waves of frequency `300 MHz` to `3000 MHz` belong to

2 Answers 4 Views

On a particular day, the maximum frequency reflected from the ionosphere is 8 MHz. On other day it was found to increase to 9 MHz. Then ratio of maxim

2 Answers 4 Views

If the frequency of the carrier wave is increased from 10 MHz to 100 MHz, the length of the half wave dipole antenna required for transmission

2 Answers 5 Views

Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - D
The motion of proton in magnetic field will be circular
`therefore v=2 pi rf=2xx3.14x50xx10^(-2)xx10xx10^(6)`
`=3.14 xx10^(7)ms^(-1)`

5 views Answered 2023-02-05 15:29