A) \[2.9\,cm\]
B) \[3.9\,cm\]
C) \[2.35\,cm\]
D) \[2\,cm\]
Correct Answer: D
Solution :
The radius of circular path followed by the electron while entering perpendicular to the magnetic field is given as \[r=\frac{mV}{eB}=\frac{V}{\left( \frac{e}{m} \right)B}\] ??(i) Here : \[\frac{e}{m}\] of electron \[=1.7\times {{10}^{11}}\text{ }C/kg\] \[V=6\times {{10}^{7}}m/s\] and \[B=1.5\times {{10}^{-2}}T\] Putting the given value in equation (i) \[\therefore \] \[r=\frac{6\times {{10}^{7}}}{1.7\times {{10}^{11}}\times 1.5\times {{10}^{-2}}}\] \[\Rightarrow \]\[r=2.35\times {{10}^{-2}}m=2.35cm\]You need to login to perform this action.
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