In a cyclotron, a magnetic induction of \(1.4 \mathrm{~T}\) is used to accelerate protons.
in Physics
17 views
2 Votes
2 Votes
In a cyclotron, a magnetic induction of \(1.4 \mathrm{~T}\) is used to accelerate protons. What should be the frequency of applied electric field? The mass and charge of proton are \(1.67 \times 10^{-27} \mathrm{~kg}\) and \(1.6 \times 10^{-19} \mathrm{C}\) respectively.

(a) \(2.5 \times 10^{7} \mathrm{~Hz}\)

(b) \(2.14 \times 10^{7} \mathrm{~Hz}\)

(c) \(3.5 \times 10^{7} \mathrm{~Hz}\)

(d) \(3.84 \times 10^{7} \mathrm{~Hz}\)
in Physics
by
3.8k Points

1 Answer

2 Votes
2 Votes
 
Best Answer

The correct option of this question will be (b).

Solution —

Here, $B=1.4 \mathrm{~T}, m=1.67 \times 10^{-27} \mathrm{~kg}$, $e=1.6 \times 10^{-19} \mathrm{C}$

The time required by a charged particle to complete a semicircle in a dee is

$$t=\frac{\pi m}{e B}=\frac{3.14 \times 1.67 \times 10^{-27}}{1.6 \times 10^{-19} \times 1.4}$$

$$=2.34 \times 10^{-8} \mathrm{~s} .$$

Thus, the direction of electric field should reverse after every $2.34 \times 10^{-8} \mathrm{~s}$.

The frequency of the applied electric field should be

$$f_{c}=\frac{1}{2 t}=\frac{1}{2 \times 2.34 \times 10^{-8}}$$

$$=2.14 \times 10^{7} \mathrm{~Hz} \text {. }$$

by
3.8k Points

Related Questions

Welcome To Informesia

Informesia Helps You To Prepare India's All States Board Exams (CBSE, ICSE, UP Board, BSEB, HPBOSE, RBSE, MSBSHSE) and Competitive Exam Like JEE (Main + Advanced), AIIMS, NEET, KVPY, NTSE, BITSAT, Olympiad, CLAT...etc.
For Any Query or Suggestion, Please Contact us on :
[email protected]

Connect us on Social Media