
$\mathrm{B}_C=\frac{\mu_0(2 \mathrm{i})}{2 \pi \mathrm{r}}+\frac{\mu_0 \mathrm{i}}{2 \pi(3 \mathrm{r})}=\frac{7 \mu_0 \mathrm{i}}{6 \pi \mathrm{r}}$
$\therefore \frac{\mathrm{B}_A}{\mathrm{~B}_C}=\frac{5}{7}$
$\therefore \mathrm{x}=5$


The ratio of the radii of trajectory of proton to that of $\alpha$ -particle is $2: 1 .$ The ratio of $K _{ p }: K _{\alpha}$ is :
