2. Sample Papers
  3. NEET

NEET Sample Paper NEET Sample Test Paper-30

  • question_answer
    An electron moves at right angle to a magnetic - field of \[1.5\times {{10}^{2}}\] tesia with a speed of \[6\times {{10}^{7}}m/s\]. If the specific charge of the electron is \[1.7\times {{10}^{11}}\] coulamb/kg. The radius of the circular path will be:

    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.
You will be redirected in 3 sec spinner