Question

magnetic field applied in a cyclotron is of magnitude 0.7πT. Frequency of electric field that should be applied between the dees in order to accelerate protons is

1) 5.35 x 106 Hz
2) 3.35 x 107 Hz
3) 5.35 x 107 Hz
4) 3.35 x 106 Hz​

Answer 1

Answer:

5.35×106Hz ans. (1) ok thank you

Answer 2

Given:

Magnetic field applied in a cyclotron has a magnitude of 0.7π Tesla.

To find:

Frequencyof electric field that should be applied in between the dees so as to accelerate protons.

Calculation:

General expression for frequency in a cyclotron is as follows:

$\boxed{ \rm{ \nu = \dfrac{qB}{2\pi m} }}$

Putting available values in SI unit:

$= > \rm{ \nu = \dfrac{q \times (0.7\pi)}{2\pi m} }$

$= > \rm{ \nu = \dfrac{q \times 0.7}{2 m} }$

$= > \rm{ \nu = \dfrac{ 0.7}{2 } \times \dfrac{q}{m} }$

$= > \rm{ \nu = \dfrac{ 0.7}{2 } \times 9.6 \times {10}^{7} }$

$= > \rm{ \nu =0.7 \times 4.8 \times {10}^{7} }$

$= > \rm{ \nu =3.36 \times {10}^{7} }$

$= > \rm{ \nu \approx3.35 \times {10}^{7} \: hz }$

So, final answer is:

$\boxed{ \bf{ \nu \approx3.35 \times {10}^{7} \: hz }}$