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AMU Medical AMU Solved Paper-2001

  • question_answer
    A 100 turns coil kept in a magnetic field  \[\overset{\to }{\mathop{\mathbf{B}}}\,=\]\[0.5\text{Wb}/{{\text{m}}^{\text{2}}},\] carries a current of 1 A. The torque acting on the coil is

    A)  1.125 N-m                         

    B)  5.25 N-m

    C)  6.25 N-m            

    D)  10 N-m

    Correct Answer: A

    Solution :

                     The torque acting is given by                 \[\tau =NI\,AB\,\sin \theta \] Given,   N = 100,  \[I=1\,A\],                 A = 15 cm \[\times \] 15 cm                 \[=225\times {{10}^{-4}}{{m}^{2}}\],                 \[B=0.5\,Wb/{{m}^{2}}\],  \[\theta ={{90}^{o}}\].                 = 1.125 N-m


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