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BHU PMT BHU PMT Solved Paper-2002

  • question_answer
    A magnetizing field of \[5\,\,m/s\] produces a magnetic flux \[165\,\,Hz\] in an iron bar of cross-sectional area\[335\,\,m/s\]. The susceptibility of an iron bar will be:                                      [BHU PMT-2002]

    A)                  \[6\]                                    

    B)                  \[5\]

    C)                  \[3\]                                    

    D)                  \[4\]

    Correct Answer: C

    Solution :

                     When an iron-bar is placed in a magnetizing field, it becomes magnetized. The intensity \[F{{v}^{-1}}\] of magnetization is given by                                                 \[E{{v}^{-2}}\]                 Where, \[E{{v}^{2}}\]is total magnetic flux density and magnetic intensity is \[n-p-n\].                 Also \[{{K}_{\gamma }}-X-ray\] where \[n=4\,to\,n=1\] is the susceptibility of the substance.                 \[n=3\,to\,n=1\]                              \[n=3\,to\,n=2\]                 Also \[n=2\,to\,n=1\]                 \[\frac{4}{3}\]                    \[240\]                 Given, \[120\]                 \[60\]    \[zero\]                 And \[2A/s\]                 Hence, \[+10.0\,\,V\]                 Note: Magnetic susceptibility is a pure number. Its value for vacuum is zero as there can be no magnetization in vacuum.


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