Question

19) A proton and an alpha particle enter at right angles into a uniform magnetic field of intensity B. Calculate the radii of their paths when they enter the field with the same.
(a) momentum and
(b) Kinetic energy

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{charge}\mathrm{of}\mathrm{proton}=\mathrm{e} \\ \mathrm{charge}\mathrm{of}\mathrm{\alpha }-\mathrm{particle}=2\mathrm{e} \\ \mathrm{mass}\mathrm{of}\mathrm{proton}=1\mathrm{u} \\ \mathrm{mass}\mathrm{of}\mathrm{\alpha }-\mathrm{particle}=4\mathrm{u} \\ \mathrm{r}=\frac{\mathrm{mv}}{\mathrm{qB}} \\ \left(1\right)\mathrm{same}\mathrm{momentum},\mathrm{p}=\mathrm{mv} \\ \mathrm{r}\mathrm{\alpha }\frac{1}{\mathrm{q}} \\ {\mathrm{r}}_{\mathrm{proton}}>{\mathrm{r}}_{\mathrm{\alpha }-\mathrm{particle}} \\ \left(2\right)\mathrm{same}\mathrm{Kinetic}\mathrm{energy},\mathrm{KE}=\frac{1}{2}{\mathrm{mv}}^2 \\ {\mathrm{r}}^2\mathrm{\alpha }\frac{{\mathrm{m}}^2{\mathrm{v}}^2}{{\mathrm{q}}^2} \\ \mathrm{Raidus}\mathrm{is}\mathrm{independent}\mathrm{of}\mathrm{kinetic}\mathrm{energy} \end{gathered}$$