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12th Class Physics Magnetism Question Bank MCQ - Magnetism and Matter

  • question_answer
       At a given place on Earth's surface the horizontal component of Earth*s magnetic field is \[2\times {{10}^{-5}}T\]and resultant magnetic field is \[4\times {{10}^{-5}}T\]. The angle of dip at this place is:

    A) \[30{}^\circ \]                          

    B)             \[60{}^\circ \]       

    C)             \[90{}^\circ \]                          

    D)             \[45{}^\circ \]       

    Correct Answer: B

    Solution :

    (b) \[60{}^\circ \] Since    \[{{B}_{H}}=B\,\cos \delta \] Here,     \[B=4\times {{10}^{-5}}T,{{B}_{H}}=2\times {{10}^{-5}}T\] \[\therefore \] \[\cos \delta =\frac{{{B}_{H}}}{B}=\frac{2\times {{10}^{-5}}}{4\times {{10}^{-5}}}=\frac{1}{2}=\cos 60{}^\circ \Rightarrow \delta =60{}^\circ \]


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