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JEE Main & Advanced AIEEE Solved Paper-2003

  • question_answer
    A particle of mass M and charge Q moving with velocity 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     AIEEE  Solved  Paper-2003

    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  \[P=F.\,v=Fv\cos \theta \]
    Here,       \[\theta ={{90}^{o}}\]
    \[\therefore \] P = 0
    But,       \[P=\frac{W}{t}\Rightarrow W=P.\,\,t\]
    Hence, work done                     W = 0          (everywhere)


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