A particle having the same charge as of electron moves in a circular path of radius $0.5
\,cm$ under the influence of a magnetic field of $0.5\,T.$ If an electric field of $100\,V/m$ makes it to move in a straight path, then the mass of the particle is (given charge of electron $= 1.6 \times 10^{-19}\, C$ )
\,cm$ under the influence of a magnetic field of $0.5\,T.$ If an electric field of $100\,V/m$ makes it to move in a straight path, then the mass of the particle is (given charge of electron $= 1.6 \times 10^{-19}\, C$ )
JEE MAIN 2019, Medium
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$\mathrm{eE}=\mathrm{evB}$
$\Rightarrow E=\left(\frac{e B r}{m}\right) B$
$\Rightarrow m=\frac{e B^{2} r}{E}$
$\Rightarrow \mathrm{m}=\frac{\left(1.6 \times 10^{-10}\right)(0.5)^{2}\left(0.5 \times 10^{-2}\right)}{100}=2 \times 10^{-24}\, \mathrm{kg}$
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