CET Karnataka Medical CET - Karnataka Medical Solved Paper-2008

  • question_answer
    The magnetic field at the centre of a circular current carrying conductor of radius r is B. The magnetic field on its axis at a distance r from the centre is Bg. The value of \[{{B}_{c}}:\text{ }{{B}_{a}}\]will be

    A)  1 : \[\sqrt{2}\]          

    B)  1 : 2\[\sqrt{2}\]

    C)  2\[\sqrt{2}\] : 1         

    D)  \[\sqrt{2}\] : 1

    Correct Answer: C

    Solution :

    Magnetic induction at the centre of the coil of radius r is \[{{B}_{c}}=\frac{{{\mu }_{0}}nI}{2r}\]                       ...(i) Magnetic induction on the axial line of a circular coil at a distance x from the centre is \[{{B}_{a}}=\frac{{{\mu }_{0}}n\,{{r}^{2}}I}{2{{({{r}^{2}}\times {{x}^{2}})}^{3/2}}}\] Given x = r \[\therefore \] \[{{B}_{a}}=\frac{{{\mu }_{0}}n\,{{r}^{2}}I}{2{{(2{{r}^{2}})}^{3/2}}}\] ?(ii) From Eqs. (i) and (ii), we get \[\frac{{{B}_{c}}}{{{B}_{a}}}=\frac{2\sqrt{2}}{1}\]


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