$\overrightarrow{\mathrm{v}} =\left(2 \times 10^{5} \hat{\mathrm{i}}\right) \mathrm{m} / \mathrm{s} $
$\overrightarrow{\mathrm{B}} =(\hat{\mathrm{i}}-4 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}) \mathrm{T} $
$\overrightarrow{\mathrm{F}} =\mathrm{e}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})$
$=\mathrm{e}\left[\left(2 \times 10^{5} \hat{\mathrm{i}}\right) \times(\hat{\mathrm{i}}-4 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})\right] $
$=2 \times 10^{5} \times 1.6 \times 10^{-19}[-4 \hat{\mathrm{k}}+3 \hat{\mathrm{j}}]$
Its magnitude is
$\mathrm{F}=2 \times 10^{5} \times 1.6 \times 10^{-19} \times \sqrt{(-4)^{2}+(3)^{2}}$
${=2 \times 10^{5} \times 1.6 \times 10^{-19} \times 5} $
${=1.6 \times 10^{-13} \mathrm{\,N}}$

$(A)$ In the region $0 < r < R$, the magnetic field is non-zero.
$(B)$ In the region $R < r < 2 R$, the magnetic field is along the common axis.
$(C)$ In the region $R < r < 2 R$, the magnetic field is tangential to the circle of radius $r$, centered on the axis.
$(D)$ In the region $r > 2 R$, the magnetic field is non-zero.

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 :