2. Solved Papers
  3. MGIMS WARDHA

MGIMS WARDHA MGIMS WARDHA Solved Paper-2004

  • question_answer
    An electron moves at right angle to a  magnetic field of \[1.5\times {{10}^{-2}}\] tesla with a speed of \[6\times {{10}^{7}}\] m/s. If the specific charge of the electron is \[1.7\times {{10}^{11}}\] coulomb/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: C

    Solution :

    The formula for radius of circular path is \[r=\frac{m\upsilon }{eB}=\frac{\upsilon }{\left( \frac{e}{m} \right)B}\]                 ?.(i) Given: \[e/m\] of electron \[=1.7\times {{10}^{11}}C/kg,\] \[\upsilon =6\times {{10}^{7}}m/s\] and        \[B=1.5\times {{10}^{-2}}T\]                               \[\therefore \]  \[r=\frac{6\times {{10}^{7}}}{1.7\times {{10}^{11}}\times 1.5\times {{10}^{-2}}}\]                             \[=2.35\times {{10}^{-2}}m\]                 \[=2.35\text{ }cm\]


You need to login to perform this action.
You will be redirected in 3 sec spinner