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

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

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

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

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