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Uttarakhand PMT Uttarakhand PMT Solved Paper-2010

  • question_answer
    A coil of area\[5c{{m}^{2}}\]and of 20 turns is placed in uniform magnetic field of\[{{10}^{3}}T\]The normal to the plane of the coil makes an angle of\[60{}^\circ \]with the magnetic field. The flux in max well through the coil, is

    A)  \[{{10}^{5}}\]

    B)  \[5\times {{10}^{4}}\]

    C)  \[2\times {{10}^{4}}\]

    D)  \[5\times {{10}^{3}}\]

    Correct Answer: B

    Solution :

     Magnetic flux \[{{\phi }_{B}}=NBA\,\cos \theta \] \[=20\times {{10}^{3}}\times 5\times cos\text{ }60{}^\circ \] \[=20\times {{10}^{3}}\times 5\times \frac{1}{2}\] \[=50\times {{10}^{3}}maxwell\] \[=5\times {{10}^{4}}maxwell\]


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