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AMU Medical AMU Solved Paper-2012

  • question_answer
    As shown in the figure, a metal rod makes contact with a partial circuit and completes the circuit. The circuit area is perpendicular to a magnetic field with B = 0.15 T. If the resistance of the total circuit is \[3\Omega \], the force needed to move the rod as indicated with a constant speed of 2 m/s will be equal to

    A)  \[3.75\times {{10}^{-3}}N\]

    B)  \[2.75\times {{10}^{-3}}N\]

    C)  \[6.57\times {{10}^{-4}}N\]       

    D)  \[4.36\times {{10}^{-4}}N\]

    Correct Answer: A

    Solution :

                     Given, \[B=0.15\,T,\,l=50\times {{10}^{-2}}cm\] and        v = 2 m/s emf \[=Bvl\]    \[e=0.15\times 2\times 50\times {{10}^{-2}}\] Current, \[i=\frac{e}{R}\]                 \[=\frac{0.15}{3}=5\times {{10}^{-2}}\] Force \[F=Bil\]                 \[=0.15\times 5\times {{10}^{-2}}\times 50\times {{10}^{-2}}\]                 \[=37.5\times {{10}^{-4}}\]                 \[=3.75\times {{10}^{-3}}N\]      


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