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JCECE Engineering JCECE Engineering Solved Paper-2002

  • question_answer
     A circular coil of radius \[4\,\,cm\] and number of turns \[20\] carries a current of \[3\,\,A\]. It is placed in a magnetic field of \[0.5\,\,T\]. The magnetic dipole moment of the coil is:

    A) \[0.60\,\,A{{m}^{2}}\]                   

    B) \[0.45\,\,A{{m}^{2}}\]

    C) \[0.30\,\,A{{m}^{2}}\]                   

    D)  \[0.15\,\,A{{m}^{2}}\]

    Correct Answer: C

    Solution :

    Magnetic dipole moment of a current carrying circular coil is given by                 \[M=Ni\,\,A=N\,\,i\,\,\pi \,\,{{r}^{2}}\] Given,\[N=20,\,\,i=3A,\,\,r=4\,\,cm=0.04\,\,m\] \[\therefore \]  \[M=20\times 3\times 3.14\times {{(0.04)}^{2}}\] \[\Rightarrow \]               \[M=0.3\,\,A-{{m}^{2}}\]


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