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12th Class Physics Magnetism Question Bank MCQ - Moving Charges and Magnetism

  • question_answer
    A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is B. It is then bent into a circular loop of smaller radius, having ‘n' turns. The magnetic field at the centre of the coil will be:

    A) \[nB\]                           

    B) \[{{n}^{2}}B\]

    C) \[2nB\]                         

    D) \[2{{n}^{2}}B\]

    Correct Answer: B

    Solution :

    (b) \[{{n}^{2}}B\]. Magnetic field at the centre of this wire.                                \[B=\frac{{{\mu }_{0}}l}{2R}\]                         …(1) The same wire is bent into a circular loop of 'n' turns.       New radius is given by.             \[n\times 2\pi r=2\pi R\Rightarrow r=\frac{R}{n}\] Thus. magnetic field at the centre of this loop \[=\frac{{{\mu }_{0}}nl}{2r}\]             \[=\frac{{{\mu }_{0}}nl}{2r}\times n={{n}^{2}}\frac{{{\mu }_{0}}l}{2R}={{n}^{2}}B\]         (from(1))


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