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JEE Main & Advanced Sample Paper JEE Main Sample Paper-30

  • question_answer
    Two consider coils X and Y have equal number of turns and carry equal currents in the same sense and subtend same solid angle at point 0. The smaller coil \[X\] is midway between 0 and Y. the magnetic induction due to bigger coil Y at 0 is \[{{B}_{Y}}\] and that due to smaller coil \[X\] at 0 is \[{{B}_{X}}\]. Then            

    A)  \[\frac{{{B}_{y}}}{{{B}_{X}}}=1\]                    

    B)  \[\frac{{{B}_{y}}}{{{B}_{X}}}=2\]

    C)  \[\frac{{{B}_{y}}}{{{B}_{X}}}=\frac{1}{2}\]   

    D) \[\frac{{{B}_{y}}}{{{B}_{X}}}=\frac{1}{4}\]

    Correct Answer: C

    Solution :

    If R is the radius of coil X, men the radius of coil Y is 2R. The magnetic fields at O due to X and Y are respectively. \[{{B}_{X}}=\frac{{{\mu }_{0}}}{2}\frac{I{{R}^{2}}}{{{({{R}^{2}}+{{d}^{2}})}^{3/2}}}\] and \[{{B}_{Y}}=\frac{{{\mu }_{0}}}{2}\frac{I{{(2R)}^{2}}}{{{[(2{{R}^{2}}+{{(2d)}^{2}})]}^{3/2}}}\] \[\therefore \frac{{{B}_{Y}}}{{{B}_{X}}}=\frac{{{2}^{2}}}{{{({{2}^{2}})}^{3/2}}}=\frac{1}{2}\]           


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