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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2004

  • 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 the magnetic meridian, a neutral point is observed at the centre of the coil. If\[{{B}_{H}}=0.314\times {{10}^{-4}}T\], the current in the coil is:

    A)  0.5 A                                    

    B)  0.25 A

    C)  2 A                                        

    D)  1 A

    Correct Answer: A

    Solution :

    At the neutral point, the magnetic field produced by the magnet becomes equal and opposite to some other magnetic field. This other magnetic field is actually the horizontal component of earths magnetic field. Magnetic field at the centre of circular coil                 \[{{B}_{H}}=\frac{{{\mu }_{0}}}{4\,\pi }\frac{2\pi \,nI}{r}\] \[I\] and r being the current and radius of circular coil respectively. or            \[I=\frac{4\,\pi }{{{\mu }_{0}}}\frac{r{{B}_{H}}}{2\pi n}\]                 \[=\frac{{{10}^{7}}\times 0.1\times 0.314\times {{10}^{-4}}}{2\times 314\times 10}\]                 \[=0.5\,A\]


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