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An electron having momentum 2.4 xx 10^(-...

An electron having momentum `2.4 xx 10^(-23)" kg m s"^(-1)` enters a region of uniform 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

A

2mm

B

1mm

C

`sqrt3/2` mm

D

0.5mm

Text Solution

Verified by Experts

The correct Answer is:
D
The radius of the helical path of the electron in the uniform magnetic field is
`r=(mv_(_|_))/(eB)=("mv sin "theta)/(eB)=((2.4xx10^(-23)"kg m s"^(-1))xxsin 30^(@))/((1.6xx10^(-19)C)xx(0.15T))`
`=5xx10^(-4)m=0.5xx10^(-3)m=0.5mm`
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