JEE Main & Advanced Sample Paper JEE Main Sample Paper-38

  • question_answer Two circular coils X and Y, having equal number of turns, carry equal currents in the same sense and subtend same solid angle at point O. If the smaller coil X is midway between O and Y, then if we represent the magnetic induction due to bigger coil Y at O as By and due to smaller coil X at O as \[{{B}_{X}}\] 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 :

     Magnetic induction at 0 due to coil Y is given by,\[{{B}_{Y}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2\pi I{{(2r)}^{2}}}{{{\left[ {{(2r)}^{2}}{{d}^{2}} \right]}^{3/2}}}\]                    ??(1) Similarly, the magnetic induction at 0 due to coil X is given by         \[{{B}_{X}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2\pi I{{r}^{2}}}{{{\left[ {{r}^{2}}+{{(d/2)}^{2}} \right]}^{3/2}}}\]     ??(2) From eq.(1) & (2)     \[\frac{{{B}_{Y}}}{{{B}_{X}}}=\frac{1}{2}\]


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