Question

A cyclotron with two cylindrical dees $($shapes of D $)$, as shown in the figure, is used to accelerate proton of mass $m=1.6\times {10}^{-27}kg$ and charge $1.6\times {10}^{-19}C$ by applying a potential difference of $100kV$ between the dees. A magnetic field of $1T$ is applied normal to the plane of dees of maximum radius of $50cm$. The number of revolutions made by the protons just before coming out of the cyclotron will be about whenever a protons leaves one of the dees , other will kept at proper polarity, is

• A. $36$
• B. $24$
• C. $42$
• D. $63$

Answer 1

The correct option is D $63$

Equating the centripetal force to the magnetic force $\frac{m{v}^2}{R}=qvB$
$\therefore R=\frac{mv}{qB}$ or $p=mv=qBR$
$(KE{)}_{max}=\frac{p^2}{2m}=\frac{(qBR{)}^2}{2m}$
In one revolution, the particle is accelerated twice along the gap between the Ds.
Energy acquired in each revolution is $2q{V}_{pot}$ where $V_{pot}$ is the potential difference applied to the Ds

No of rotations: $n=\frac{(KE{)}_{max}}{2q{V}_{pot}}=\frac{(qBR{)}^2}{2m\times 2q{V}_{pot}}=\frac{q}{m}\frac{B^2{R}^2}{4V_{pot}}=\frac{1.6\times {10}^{-19}}{1.6\times {10}^{-27}}\times \frac{1^2\times 1^2}{4\times 4\times {10}^5}=63revs$