JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2007

  • question_answer
        The magnetic field due to a current carrying circular loop of radius 3 cm at a point on the axis at a distance of 4 cm from the centre is\[5\mu T\]. What will be its value at the centre of the loop?

    A)  \[250\mu T\]                    

    B)  \[150\mu T\]

    C)  \[125\mu T\]                    

    D)  \[75\mu T\]

    Correct Answer: A

    Solution :

                    The magnetic field at a point on the axis of a circular loop at a distance\[x\]from the centre is \[B=\frac{{{\mu }_{0}}i{{R}^{2}}}{2{{({{R}^{2}}+{{x}^{2}})}^{3/2}}}\]                         ?(i) Given: \[B=54\text{ }\mu T,\text{ }x=4\text{ }cm,\text{ }R=3\text{ }cm\] Putting the given values in Eq. (i), we get \[\therefore \]  \[54=\frac{{{\mu }_{0}}i\times {{(3)}^{2}}}{2{{({{3}^{2}}+{{4}^{2}})}^{3/2}}}\] \[\Rightarrow \]               \[54=\frac{9{{\mu }_{0}}i}{2{{(25)}^{3/2}}}=\frac{9{{\mu }_{0}}i}{2\times {{(5)}^{3}}}\] \[\therefore \]  \[{{\mu }_{0}}i=\frac{54\times 2\times 125}{9}\]                 \[{{\mu }_{0}}i=1500\,\mu T-cm\]                            ?(ii) Now, putting\[x=0\]in Eq. (i), magnetic field at the centre of loop is                 \[B=\frac{{{\mu }_{0}}i{{R}^{2}}}{2{{R}^{3}}}=\frac{{{\mu }_{0}}i}{2R}=\frac{1500}{2\times 3}\] \[=250\text{ }\mu T\]                  [from Eq. (ii)]


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