Question

The magnetic field applied in a cyclotron is of magnitude 0.7pie T. Frequency of elctric field that should be applied between the dees in order to accelerate proton,is____.​

Answer 1

Given:

The magnetic field applied in a cyclotron is of magnitude 0.7pie T.

To find:

Frequency of elctric field that should be applied between the dees in order to accelerate proton.

Calculation:

Time period in a cyclotron $(\tau)$

$\therefore \: \tau = \dfrac{2\pi r}{v}$

$= > \: \tau = \dfrac{2\pi ( \frac{mv}{Bq}) }{v}$

$= > \: \tau = \dfrac{2\pi ( \frac{m \cancel{v}}{Bq}) }{ \cancel{v}}$

$= > \: \tau = \dfrac{2\pi m}{Bq}$

So, frequency is reciprocal of Time period:

$= > \: f = \dfrac{1}{ \tau}$

$= > \: f = \dfrac{Bq}{ 2\pi m}$

$= > \: f = \dfrac{0.7\pi }{ 2\pi } \times \dfrac{q}{m}$

$= > \: f = \dfrac{0.7}{ 2 } \times 1.04 \times {10}^{ - 8}$

$= > \: f = 0.35 \times 1.04 \times {10}^{ - 8}$

$= > \: f = 0.364 \times {10}^{ - 8} \: hz$

So, final answer is:

$\boxed{ \sf{\: f = 0.364 \times {10}^{ - 8} \: hz}}$