4.Moving Charges and Magnetism
medium

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

A

$9.1 \times 10^{-31}\, kg$ 

B

$1.6 \times 10^{-27}\, kg$ 

C

$1.6 \times 10^{-19}\, kg$ 

D

$2.0 \times 10^{-24}\, kg$ 

(JEE MAIN-2019)

Solution

$\mathrm{eE}=\mathrm{evB}$

$\Rightarrow  E=\left(\frac{e B r}{m}\right) B$

$\Rightarrow  m=\frac{e B^{2} r}{E}$

$\Rightarrow  \mathrm{m}=\frac{\left(1.6 \times 10^{-10}\right)(0.5)^{2}\left(0.5 \times 10^{-2}\right)}{100}=2 \times 10^{-24}\, \mathrm{kg}$

Standard 12
Physics

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