2. Solved Papers
  3. MGIMS WARDHA

MGIMS WARDHA MGIMS WARDHA Solved Paper-2010

  • question_answer
    A particle of mass M and charge \[Q\] moving with velocity \[\mathbf{\vec{V}}\] describes a circular path of radius R when subjected to a uniform transverse magnetic field of induction B. The work done by the field when the particle completes one full circle is

    A)  \[\left( \frac{M{{v}^{2}}}{R} \right)2\pi R\]         

    B)  zero

    C)  \[BQ\,\,2\pi R\]                              

    D)  \[BQv\,\,2\pi R\]

    Correct Answer: B

    Solution :

                     When particle describes circular path in a magnetic field, its velocity is always perpendicular to the magnetic force. Power,   \[P=\overrightarrow{F}-\overrightarrow{v}=Fv\cos \theta \] Here, \[\theta =90{}^\circ \] \[\therefore \]  \[P=0\] But          \[P=\frac{W}{t}\] \[\Rightarrow \]               \[W=P.t\] Hence, work done \[W=0\]                                       (Everywhere)


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