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 beA. 2mmB. 1mmC. `sqrt3/2` mmD. 0.5mm

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

1.	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 beA. 2mmB. 1mmC. `sqrt3/2` mmD. 0.5mm
Answer» Correct Answer - 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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