CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2008

  • question_answer
    The magnetic field at the centre of a circular current carrying conductor of radius r is \[{{B}_{c}}\]. The magnetic field on its axis at a distance r from the centre is \[{{B}_{a}}\]. The value of \[{{B}_{c}}:{{B}_{a}}\]  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                 \[{{B}_{c}}=\frac{{{\mu }_{0}}nI}{2\,r}\]                                ... (i) Magnetic induction on the axial line of a circular coil at a distance x from the centre is                 \[{{B}_{a}}=\frac{{{\mu }_{0}}n{{r}^{2}}I}{2{{({{r}^{2}}+{{x}^{2}})}^{3/2}}}\] 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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