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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2000

  • question_answer
    A solenoid of length 50 cm and a radius of cross-section 1 cm has 1000 turns of wire wound over it. If the current carried is 5 A, the magnetic field on its axis, near the centre of the solenoid is approximately (permeability of free space\[{{\mu }_{0}}=4\pi \times {{10}^{-7}}T-m/A)\]:

    A)  \[0.63\times {{10}^{-2}}T\]        

    B)  \[1.26\times {{10}^{-2}}T\]        

    C)  \[2.51\times {{10}^{-2}}T\]

    D)         \[6.3T\]

    E)  \[12.6\text{ }T\]

    Correct Answer: B

    Solution :

    The magnetic field is given by \[B={{\mu }_{0}}ni\] where \[{{\mu }_{0}}=4\pi \times {{10}^{-7}}T\,m{{A}^{-1}},\]                 \[n=\frac{1000}{50\times {{10}^{-2}}},i=5A\] \[\therefore \]  \[B=4\pi \times {{10}^{-7}}\times \frac{1000\times {{10}^{-2}}}{50\times {{10}^{-2}}}\times 5\]                 \[B=1.26\times {{10}^{-2}}T\]


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