An electron moves with a speed of 2 times 10^ | vedclass

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Question

Given

$\overrightarrow{\mathrm{v}} =\left(2 	imes 10^{5} \hat{\mathrm{i}}ight) \mathrm{m} / \mathrm{s} $

$\overrightarrow{\mathrm{B}} =(\hat{

Vedclass · Accepted Answer

$1.6 	imes {10^{ - 13}}$

Vedclass · Answer

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

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 $= 1.6 \times 10^{-19}\, C$)

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