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WB JEE Medical WB JEE Medical Solved Paper-2006

  • question_answer
    The coil in a tangent galvanometer is 16 cm in radius. If a current of 20 mA is to produce a deflection of \[\text{4}{{\text{5}}^{\text{o}}},\] then the number of turns wound on it, is (Take horizontal component of earths magnetic field \[=0.36\times {{10}^{-4}}T\]and\[{{\mu }_{0}}=4\pi \times {{10}^{-7}}\,Wb{{A}^{-1}}{{m}^{-1}}\])

    A)  229                                       

    B)  458

    C)  689                                       

    D)  916

    Correct Answer: B

    Solution :

                     Current in coil of tangent galvanometer \[i=\frac{2rH}{{{\mu }_{0}}n}\tan \theta \]                 \[\Rightarrow \]               \[n=\frac{2rH}{{{\mu }_{0}}i}\tan \theta \]                 \[\therefore \]  \[n=\frac{2\times 16\times {{10}^{-2}}\times 0.36\times {{10}^{-4}}}{4\pi \times {{10}^{-7}}\times 20\times {{10}^{-3}}}\tan {{45}^{o}}\] \[=458\]


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