A charge of $4.0 \mu \mathrm{C}$ is moving with a velocity of $4.0 \times 10^6 \mathrm{~ms}^{-1}$ along the positive $\mathrm{y}$-axis under a magnetic field $\vec{B}$ of strength $(2 \hat{k})\ T$. The force acting on the charge is $x \hat{i} N$. The value of $x$ is__________.
JEE MAIN 2024, Diffcult
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$\mathrm{q} =4 \mu \mathrm{C}, \overrightarrow{\mathrm{v}}=4 \times 10^6$
$ \hat{\mathrm{j}} \mathrm{m} / \mathrm{s} $
$\overrightarrow{\mathrm{B}} =2 \hat{\mathrm{k}} \mathrm{T} $
$\overrightarrow{\mathrm{F}} =\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}}) $
$ =4 \times 10^{-6}\left(4 \times 10^6 \hat{\mathrm{j}} \times 2 \hat{\mathrm{k}}\right) $
$ =4 \times 10^{-6} \times 8 \times 10^6 \hat{\mathrm{i}} $
$\overrightarrow{\mathrm{F}} =32 \hat{\mathrm{i}} \mathrm{N} $
$\mathrm{x} =32$
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