Question

A particle of mass 'm' and charge 'Q' is acceleration in cyclotron. if magnetic field is 'B' and radius cyclotron is r then the kinetic energy of particle.

• A. $\frac{QBr}{2m}$
• B. $\frac{Q^2{B}^{2}r}{2m}$
• C. $\frac{Q^2{B}^2{r}^2}{2m}$
• D. $\frac{Q^2{B}^2{r}^2}{m}$

Answer 1

The correct option is B $\frac{Q^2{B}^{2}r}{2m}$

Mass $=m$

charge $=Q$

magnitude field $=B$

radius cyclotron $=r$

Kinetic energy of particle $\frac{1}{2}{mV}^2=K.E$

where $V=$ velocity of cyclotron.

Calculation of the cyclotron frequency (non-realistic). The Lorentz force

$F_{lorentz}$ is the centripetal force $F_{Zcentripetal}$ and cause the particles path to bend in a circle.

$F_{lorentz}=F_{Zcentripetal}$

$\Rightarrow q.V.B=\frac{{mV}^2}{r}$

$V=\frac{r.q.B}{m}$

now,

$V=w.r$

$=2\pi f.r$

The cyclotron frequency

$K.E=\frac{1}{2}m\frac{r^2{q}^2{B}^2}{m^2}$

$K.E=\frac{1}{2}\frac{Q^2{B}^{2}r}{m}$