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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2001

  • question_answer
    The magnetic moment of current loop is\[2.1\times {{10}^{-25}}\]\[A-{{m}^{2}}\]. The magnetic field at distance 1 \[\overset{o}{\mathop{\text{A}}}\,\]from centre on axis of loop will be

    A)  \[4.2\times {{10}^{2}}Wb/{{m}^{2}}\]

    B)  \[4.2\times {{10}^{-3}}Wb/{{m}^{2}}\]

    C)  \[4.2\times {{10}^{-4~}}Wb/{{m}^{2}}\]

    D)  \[4.2\times {{10}^{-5}}Wb/{{m}^{2}}\]

    Correct Answer: A

    Solution :

     Intensity of field at a distance x from centre of loop \[F=\frac{{{\mu }_{0}}}{4\pi }\frac{2M}{{{x}^{3}}}=\frac{{{10}^{-7}}\times 2\times 2.1\times {{10}^{-25}}}{{{({{10}^{-10}})}^{3}}}\] \[=4.2\times {{10}^{-2}}Wb/{{m}^{2}}\]


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