JEE MAIN - Physics (2024 - 29th January Evening Shift - No. 29)
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 $$y$$ axis under a magnetic field $$\vec{B}$$ of strength $$(2 \hat{k}) \mathrm{T}$$. The force acting on the charge is $$x \hat{i} N$$. The value of $$x$$ is __________.
Answer
32
Explanation
$$\begin{aligned}
\mathrm{q} & =4 \mu \mathrm{C}, \overrightarrow{\mathrm{v}}=4 \times 10^6 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s} \\
\overrightarrow{\mathrm{B}} & =2 \hat{\mathrm{k} T} \\
\overrightarrow{\mathrm{F}} & =\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}}) \\
& =4 \times 10^{-6}\left(4 \times 10^6 \hat{\mathrm{j}} \times 2 \hat{\mathrm{k}}\right) \\
& =4 \times 10^{-6} \times 8 \times 10^6 \hat{\mathrm{i}} \\
\overrightarrow{\mathrm{F}} & =32 \hat{\mathrm{i}} \mathrm{N} \\
\mathrm{x} & =32
\end{aligned}$$
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