WB JEE Medical WB JEE Medical Solved Paper-2013

  • question_answer
    When a proton is released from rest in a room, it starts with an initial acceleration\[{{a}_{0}}\] towards west. When it is projected towards north with a speed \[{{\upsilon }_{0}}\]it moves with an initial acceleration \[3{{a}_{0}}\] towards west. The electric and magnetic fields in the room are

    A) \[\frac{m{{a}_{0}}}{e}west,\frac{2m{{a}_{0}}}{e{{v}_{0}}}up\]

    B)  \[\frac{m{{a}_{0}}}{e}west,\frac{2m{{a}_{0}}}{e{{v}_{0}}}down\]

    C)  \[\frac{m{{a}_{0}}}{e}east,\frac{3m{{a}_{0}}}{e{{v}_{0}}}up\]

    D)  \[\frac{m{{a}_{0}}}{e}east,\frac{3m{{a}_{0}}}{e{{v}_{0}}}down\]

    Correct Answer: B

    Solution :

     Initial acceleration, \[{{a}_{0}}=\frac{eE}{m}\] ?(i) \[\Rightarrow \] \[E=\frac{{{a}_{0}}m}{e}\]  \[\therefore \]  \[\frac{e{{v}_{0}}B+eE}{m}=3{{a}_{0}}\] or \[e{{v}_{0}}B+eE=3{{a}_{0}}m\] \[\therefore \] \[e{{v}_{0}}B=3m{{a}_{0}}-eE\] \[\Rightarrow \] \[=3\,m{{a}_{0}}-m{{a}_{0}}\] [from eq. (i)] \[\Rightarrow \] \[e{{v}_{0}}B=2m{{a}_{0}}\] \[\therefore \] \[B=\frac{2m{{a}_{0}}}{e{{v}_{0}}}\]


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