Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2005

  • question_answer
    A current of 1 A is passed through a straight wire of length 2 m. The magnetic field at a point in air at a distance of 3 m from either end of wire and lying on the axis of wire will be

    A)  \[\frac{{{\mu }_{0}}}{2\pi }\]                                     

    B) \[\frac{{{\mu }_{0}}}{4\pi }\]

    C) \[\frac{{{\mu }_{0}}}{8\pi }\]                                      

    D)  zero

    Correct Answer: D

    Solution :

    According to Biot-Savart's law - magnetic field at distance r from current carrying conductor is \[dB=\frac{{{\mu }_{0}}}{4\pi }\,\,\frac{Idl\,\sin \theta }{{{r}^{2}}}\] For a point lying on axis of wire \[\theta =0\]. So, magnetic field at any point on the axis \[dB=0\].                          (\[\because \] \[\sin \,0=0\])


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