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 Vm makes it to move in a straight path, then the mass of the particle is given by x×10-24 kg where the value of x is ________. (Given charge of electron =1.6×10-19 C)

r=mvqB ...i Also, FB=FC or qvB=qE or rqBm×B=E ⇒ m=rqB2E [Using i]∴ m=0.5×10-2×1.6×10-19×0.52100=2×10-24 kg

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A particle having the same charge as of electron moves in a circular path of radius $0.5 c m$ under the influence of a magnetic field of $0.5 T$ . If an electric field of $100 \frac{V}{m}$ makes it to move in a straight path, then the mass of the particle is given by $x \times 10^{- 24} k g$ where the value of $x$ is $________.$ (Given charge of electron $= 1.6 \times 10^{- 19} C$ )

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Detailed Solution

$r = \frac{m v}{q B} . . . (i)$

Also, $F_{B} = F_{C}$ or $q v B = q E$

or $\frac{r q B}{m} \times B = E \Rightarrow m = \frac{r q B^{2}}{E}$ [Using $(i)$ ]

$∴ m = \frac{0.5 \times 10^{- 2} \times 1.6 \times 10^{- 19} \times {(0.5)}^{2}}{100} = 2 \times 10^{- 24} k g$

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