In a cyclotron, a magnetic induction of 1.4 T is used to accelerate protons.

Question

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) \(3.84 \times 10^{7} \mathrm{~Hz}\)

Answer 1

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

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

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