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An electron moves with a speed of $2 \times 10^5\, m/s$ along the $+ x$ direction in a magnetic field $\vec B = \left( {\hat i - 4\hat j - 3\hat k} \right)\,tesla$. The magnitude of the force (in newton) experienced by the electron is (the charge on electron $= 1.6 \times 10^{-19}\, C$)
$1.18 \times {10^{ - 13}}$
$1.28 \times {10^{ - 13}}$
$1.6 \times {10^{ - 13}}$
$1.72 \times {10^{ - 13}}$
Solution
Given
$\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} $
${=1.6 \times 10^{-13} \mathrm{\,N}}$
