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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2008

  • 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   \[P=\overset{\to }{\mathop{\mathbf{F}}}\,\cdot \overset{\to }{\mathop{\mathbf{v}}}\,=Fv\cos \theta \] Here,     \[\theta ={{90}^{o}}\]                 \[P=0\] But         \[P=\frac{W}{t}\] \[\Rightarrow \]               \[W=P\cdot t\] Hence, work done \[W=0\] (everywhere)


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