2. Solved Papers
  3. VIT Engineering

VIT Engineering VIT Engineering Solved Paper-2015

  • question_answer
    An electron of mass \[9.0\times {{10}^{-31}}\]kg under the action of a magnetic field moves in a circle of radius 2 cm at a speed of \[3\times {{10}^{6}}m/s.\]If a proton of mass \[1.8\times {{10}^{27}}\]kg was to move in a circle of same radius in the same magnetic field, then its speed will become

    A) \[1.5\times {{10}^{3}}m/s\]       

    B) \[3\times {{10}^{6}}m/s\]

    C) \[6\times {{10}^{4}}m/s\]           

    D) \[2\times {{10}^{6}}m/s\]    

    Correct Answer: A

    Solution :

    Here, the magnetic force will provide the necessary centripetal force \[\therefore \]              \[\left| \frac{\Delta p}{p} \right|=\left| \frac{\Delta \lambda }{\lambda } \right|\]                    Bqr  =  mv For electron and proton, the magnetic field B, charge q and radius r, all same. So,        mv  = constant i.e.       \[\Rightarrow \]              \[\frac{p'}{p}=\frac{0.20}{100}=\frac{1}{500}\]              \[\Rightarrow \]


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