Question

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^(-4)m`

Answer 1

Correct Answer - 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`