The figure shows a region of length ‘l’ with a uniform magnetic field of 0.3 T in it and a proton entering the region with velocity `4xx10^(5) " ms"^(-1)` making an angle `60^(@)` with the field. If the proton completes 10 revolution by the time it cross the region shown, ‘l’ is close to (mass of proton =` 1.67 xx10^(27)` kg , charge of the proton ` =1.6 xx10^(-19) ` C )

Options

A. `0.11` m
B. `0.22`m
C. `0.44`m
D. `0.88`m

Answer

C

Explanation

` l = 10 x " pitch"`
` = 10 xx v cos 60^(@) xx (2pim)/(qB)`
` = 10 xx v cos 60^(@) xx (2pim)/(qB)`
` = 10 xx v xx 1/2 xx (2pim)/(qB)`
` l = (10 vpim)/(qB) = (10xx4xx10^(5) xx 3.14 xx1.67 xx10^(-27))/(1.6 xx10^(-19) xx0.3)`
` cong 0.44 m `