VIT Engineering VIT Engineering Solved Paper-2010

  • question_answer
    A cylindrical conductor of radius R carries a current i. The value of magnetic field at a point which is \[\frac{R}{4}\]distance inside from the surface is 10 T. The value of magnetic field at point which is 4R distance outside from the surface

    A)  \[\frac{4}{3}T\]              

    B)  \[\frac{8}{3}T\]

    C)  \[\frac{40}{3}T\]

    D)  \[\frac{80}{3}T\]  

    Correct Answer: B

    Solution :

    Magnetic field inside the cylindrical conductor \[{{B}_{in}}=\frac{{{\mu }_{0}}}{4\pi }\frac{2ir}{{{R}^{2}}}\] [R = radius opf cylinder, r = distance of observation point from axis of cylinder) Magnetic field outside the cylinder at a distance r from its axis \[{{B}_{in}}=\frac{{{\mu }_{0}}}{4\pi }\frac{2i}{r}\] \[\Rightarrow \]\[\frac{{{B}_{in}}}{{{B}_{out}}}=\frac{rr}{{{R}^{2}}}\]\[\Rightarrow \] \[\frac{10}{{{B}_{out}}}=\frac{\left[ R-\frac{R}{4} \right](R+4R)}{{{R}^{2}}}\] \[\Rightarrow \] \[{{B}_{out}}=\frac{8}{3}T\]


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