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AFMC AFMC Solved Paper-2008

  • 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 I ampere flows through the wire as shown. The magnitude of the force on PQ and QR will be

    A) BHIl, 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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