A) \[10m\]
B) \[1\text{ }m~\]
C) \[0.1m\]
D) \[100m\]
Correct Answer: D
Solution :
: When the proton beam enters the magnetic field B normally, it describes a circular path of radius r given by \[r=\frac{mv}{eB}=\frac{v}{\frac{e}{m}B}\] where \[\frac{e}{m}\] is the specific charge of the proton and v is its velocity. Here, \[v={{10}^{9}}\,m\,{{s}^{-1}},\frac{e}{m}={{10}^{11}}C\,k{{g}^{-1}}\] \[B={{10}^{-4}}\,Wb\,{{m}^{-2}}\] \[\therefore \] \[r=\frac{{{10}^{9}}\,m\,{{s}^{-1}}}{({{10}^{11}}C\,k{{g}^{-1}})({{10}^{-4}}Wb\,{{m}^{-2}})}=100m\]You need to login to perform this action.
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