CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2006

  • question_answer
    A solenoid 1.5m long and 0.4 cm in diameter possesses 10 turns per cm length. A current of 5 A falls through it. The magnetic field at the axis inside the solenoid is:

    A)  \[2\pi \times {{10}^{-3}}T\]                        

    B)  \[2\pi \times {{10}^{-5}}T\]

    C)  \[4\pi \times {{10}^{-2}}T\]                        

    D)  \[4\pi \times {{10}^{-3}}T\]

    Correct Answer: A

    Solution :

    Magnetic field at the axis inside the solenoid                                 \[B={{\mu }_{0}}\,ni\] Here, \[n=10\,\,turns/cm=1000\text{ }turns/m,\] \[\therefore \]  \[B\times 4\pi \times {{10}^{-7}}\times 1000\times 5\]                 \[=2\pi \times {{10}^{-3}}T\].


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