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CET Karnataka Medical CET - Karnataka Medical 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 = I. A current I ampere flows through the wire as shown. The magnitude of the force on PQ and QR 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 \[\theta =0,\] \[\therefore \] \[{{F}_{PQ}}=IBl\sin {{0}^{o}}=0\] Also, wire QR is perpendicular to the direction of magnetic field, then \[\theta ={{90}^{o}}.\] \[\therefore \] \[{{F}_{QR}}=IBl\sin ={{90}^{o}}=IBl\]


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