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An electron of energy 1800 eV describes ...

An electron of energy 1800 eV describes a circular path in magnetic field of flux density 0.4 T. The radius of path is `(q = 1.6 xx 10^(-19) C, m_(e)=9.1 xx 10^(-31) kg)`

A

`2.58xx10^(-4)m`

B

`3.58xx10^(-4)m`

C

`2.58xx10^(-3)m`

D

`3.58xx10^(-3)m`

Text Solution

Verified by Experts

The correct Answer is:
B
`E=(1)/(2)mv^(2) rArr v=sqrt((2E)/(m))`
`r=(mv)/(Be)=(m)/(Be)sqrt((2E)/(m))=(sqrt(2mE))/(Be)`
`r=(sqrt(2xx1800xx1.6xx10^(-19)xx9.1xx10^(-31)))/(1.6xx10^(-19)xx0.4)=3.58xx10^(-4)m`
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