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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 27

  • question_answer
    A vertical circular coil of radius 0.1 m and having 10 turns carries a steady current. When the plane of the coil is normal to magnetic meridian, a neutral point is observed at the centre of the coil. If \[{{B}_{H}}=0.314\times {{10}^{-4}}T\], then current in the coil is                          

    A) 0.25 A  

    B) 0.5 A

    C) 1 A                  

    D) 2 A     

    Correct Answer: B

    Solution :

    [b] : Given: r= 1.m, N=10, \[{{B}_{H}}=0.314\times {{10}^{-4}}T\] Magnetic field at the centre of current carrying circular coil\[B=\frac{{{\mu }_{0}}NI}{2r}\] Since at neutral point, the magnetic field due to circular coil is completely cancelled by the horizontal component of earths magnetic field. Therefore, \[B={{B}_{H}}\,\text{or}\frac{{{\mu }_{0}}NI}{2r}={{B}_{H}}\] \[\therefore \]\[I=\frac{2r{{B}_{H}}}{{{\mu }_{0}}N}=\frac{2\times 0.1\times 0.314\times {{10}^{-4}}}{(4\pi \times {{10}^{-7}})\times 10}=0.5A\]


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