Question

Maximum kinetic energy of the positive ion in the cyclotron is

• A. $\frac{q^{2}B{r}_0}{2m}$
• B. $\frac{q{B}^2{r}_0}{2m}$
• C. $\frac{q^2{B}^2{r}_0^2}{2m}$
• D. $\frac{qB{r}_0}{2m^2}$

Answer 1

The correct option is C $\frac{q^2{B}^2{r}_0^2}{2m}$

Let $r_0$ be maximum radius of circular path of positive ion.
Let $v_0$ be maximum speed of positive ion.
$q{v}_{0}B=\frac{m{v}_0^2}{r_0}$
$v_0=\frac{qB{r}_0}{m}$
Maximum kinetic energy of ion =$\frac{m{v}_0^2}{2}=\frac{1}{2}m(\frac{qB{r}_0}{m}{)}^2=\frac{B^2{q}^2{r}_0^2}{2m}$