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VMMC VMMC Medical Solved Paper-2015

  • question_answer
    Two short bar magnets with magnetic moments 400 Wb-A \[c{{m}^{2}}\]and 800 Wb-A\[c{{m}^{2}}\]are placed with their axis in the same straight line with similar poles facing each other and with their centres at 20 cm from each other. Then, the force of repulsion is

    A) 0.2T                                       

    B) 0.7T

    C) 0.4T   

    D) 0.9T

    Correct Answer: C

    Solution :

    As, magnetic moment \[M=niA=50\times 2\times 1.25\times {{10}^{-3}}=0.125A{{m}^{2}}\] If the normal to face of the coil makes an angle \[\theta \] with the magnetic induction B, then in 1st case, torque \[=MB\sin \theta =0.03\] \[\therefore \]\[MB=\sqrt{{{(0.04)}^{2}}+{{(0.03)}^{2}}}=0.05\] \[\Rightarrow \]\[B=\frac{0.05}{M}=\frac{0.05}{0.125}=0.4T\]                                


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