NEET AIPMT SOLVED PAPER MAINS 2010

  • question_answer
    A current loop consists of two identical semicircular parts each of radius R, one lying in the x-y plane and the other in x-z plane. If the current in the loop is i. The resultant magnetic field due to the two semicircular parts at their  common centre is              

    A) \[\frac{{{\mu }_{0}}i}{2\sqrt{2}R}\]         

    B) \[\frac{{{\mu }_{0}}i}{2R}\]                         

    C) \[\frac{{{\mu }_{0}}i}{4R}\]                         

    D) \[\frac{{{\mu }_{0}}i}{\sqrt{2}R}\]

    Correct Answer: A

    Solution :

    The magnetic field at centre O for each semicircular parts each of radius R, \[{{B}_{1}}={{B}_{2}}=\frac{{{\mu }_{0}}i}{4R}\] The magnetic field at their common centre \[\overset{\to }{\mathop{\mathbf{B}}}\,=\overset{\to }{\mathop{{{\mathbf{B}}_{1}}}}\,+\overset{\to }{\mathop{{{\mathbf{B}}_{2}}}}\,\] \[B=\sqrt{B_{1}^{2}+B_{2}^{2}}\] \[=\sqrt{{{\left( \frac{{{\mu }_{0}}i}{4R} \right)}^{2}}+{{\left( \frac{{{\mu }_{0}}i}{4R} \right)}^{2}}}\] \[B=\frac{{{\mu }_{0}}i}{2\sqrt{2}R}\]


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