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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2006

  • 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 \[PQ=QR=l.\text{ }A\] current \[I\]ampere flows through the wire as shown. The magnitude of the force on PQ and Q.R will be:

    A)  \[BIl,0\]                              

    B)  \[2BIl,0\]

    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 of magnetic field, then \[\theta =0\], \[\therefore \]  \[{{F}_{PQ}}=IBl\,\sin \theta \] Also, wire QR is perpendicular to the direction of magnetic field, then \[\theta ={{90}^{o}}\]. \[\therefore \]  \[{{F}_{PQ}}=IBl\sin {{0}^{o}}=0\]


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