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JCECE Engineering JCECE Engineering Solved Paper-2009

  • question_answer
    A wire \[PQR\] is bent as shown in figure and is placed in a region of uniform magnetic field\[B\]. The length of \[PQ=QR=l\,\,A\] current \[l\] ampere flows through the wire as shown. The magnitude of the force on \[PQ\] and \[QR\] will be

    A) \[BIl,\,\,0\]                        

    B) \[BIl,\,\,0\]

    C) \[0,\,\,BIl\]                        

    D) \[0,\,\,0\]

    Correct Answer: C

    Solution :

    The Lorentz force acting on the current carrying conductor in the magnetic field is                 \[F=IBl\sin \theta \] Since, wire \[PQ\] is parallel to the direction magnetic field, then\[\theta =0\], \[\therefore \]  \[{{F}_{PQ}}=IBl\sin {{0}^{o}}=0\] Also, wire \[QR\] is perpendicular to the direction magnetic field, then\[\theta ={{90}^{o}}\]. \[\therefore \]  \[{{F}_{QR}}=IBl\sin {{90}^{o}}=IBl\]


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