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A narrow beam of protons and deuterons, each having the same momentum, enters a region of uniform magnetic field directed perpendicular to their direction of momentum. The ratio of the radii of the circular paths described by them is
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Since, the radius of circular path of a charged
particle in magnetic field is $\mathrm{r}=\frac{\mathrm{mv}}{\mathrm{qB}}=\frac{\rho}{\mathrm{qB}}$
Now, the radius of circular path of charged particle of given momentum $\rho$ and magnetic field $\mathrm{B}$ is given by $\mathrm{r} \propto \frac{1}{\mathrm{q}}$
But charge on both charged particles, protons and deuterons, is same. Therefore,
$$
\frac{\mathrm{r}_{\mathrm{p}}}{\mathrm{r}_{\mathrm{D}}}=\frac{\mathrm{q}_{\mathrm{D}}}{\mathrm{q}_{\rho}}=\frac{1}{\mathrm{l}}
$$
particle in magnetic field is $\mathrm{r}=\frac{\mathrm{mv}}{\mathrm{qB}}=\frac{\rho}{\mathrm{qB}}$
Now, the radius of circular path of charged particle of given momentum $\rho$ and magnetic field $\mathrm{B}$ is given by $\mathrm{r} \propto \frac{1}{\mathrm{q}}$
But charge on both charged particles, protons and deuterons, is same. Therefore,
$$
\frac{\mathrm{r}_{\mathrm{p}}}{\mathrm{r}_{\mathrm{D}}}=\frac{\mathrm{q}_{\mathrm{D}}}{\mathrm{q}_{\rho}}=\frac{1}{\mathrm{l}}
$$
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