Question

A proton is accelerating in a cyclotron. It comes out of the cyclotron with the kinetic energy $K$. If it is required to have kinetic energy $4K$, then the radius of the circular path will be _____ times of its initial value.

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

We know that, radius of the circular path,

$r=\frac{mv}{qB}$

$\Rightarrow v=\frac{rqB}{m}$

Now, kinetic energy,

$K=\frac{1}{2}\times m\times {\left(\frac{rqB}{m}\right)}^2=\frac{r^2{q}^2{B}^2}{2m}$

For the same charged particle and cyclotron,

$r\propto \sqrt{K}$

So, if $K^{{}^{\prime }}=4K$ then $r^{{}^{\prime }}=2r$