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NEET NEET SOLVED PAPER 2016 Phase-II

  • question_answer
    A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is B. It is then bent into a circular coil of n turns. The magnetic field at the centre of this coil of n turns will be

    A)  nB                              

    B)  \[{{n}^{2}}B\]

    C)  2nB                            

    D)  \[2{{n}^{2}}B\]

    Correct Answer: B

    Solution :

                B= \[\frac{{{\mu }_{o}}I}{2r}\] when made an turn radius becomes r`             \[n\times 2\pi r\grave{\ }=2\pi r\Rightarrow r\grave{\ }=\frac{r}{n}\]             Now, B`=\[\frac{{{\mu }_{{}^\circ }}nI}{2r\grave{\ }}={{n}^{2}}\frac{{{\mu }_{{}^\circ }}I}{2r}={{n}^{2}}B\]   


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