An electron having momentum `2.4 xx 10^(-23)` kg m"/"s enters a region of unifom magnetic field of `0.15` T. The field vector makes an angle of `30^(@)` with the initial velocity Vector of the electron. The radius of the helical path of the electron in the field shall be

Options

A. 2mm
B. 1mm
C. `(sqrt3)/(2)mm`
D. 0.5mm

Answer

D

Explanation

The radius of the helical path of the electron in the uniforme magnetic field is `r=(mv_(bot))/(eB)=(mv sintheta)/(eB)=((2.4xx10^(-23)kgm"/"s)xxsin30^(@))/((1.6xx10^(-19)C)xx0.15T)`
`=5xx10^(-4)m=0.5xx10^(-6)m=0.5mm`