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NEET Sample Paper NEET Sample Test Paper-90

  • question_answer
    Consider two thin identical conducting wires covered with very thin insulating material. One of the wires is bent into a loop and produces magnetic field \[{{B}_{1}},\] at its centre when a current I passes through it. The ratio \[{{B}_{1}}:{{B}_{2}}\] is:

    A) 1 : 1                 

    B)        1 : 3 

    C) 1 : 9                 

    D)         9 : 1

    Correct Answer: B

    Solution :

    [b] For loop B\[=\frac{{{\mu }_{0}}nI}{2a}\] where, a is the radius of loop. Then, \[{{B}_{1}}=\frac{{{\mu }_{0}}I}{2a}\] Now, for coil \[B=\frac{{{\mu }_{0}}I}{4\pi }.\frac{2nA}{{{x}^{3}}}\] at the centre x = radius of loop \[{{B}_{2}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{2\times 3\times (I/3)\times \pi {{(a/3)}^{2}}}{{{(a/3)}^{3}}}\] \[=\frac{{{\mu }_{0.3}}I}{2a}\] \[\therefore \] \[\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{{{\mu }_{0}}I/2a}{{{\mu }_{0}}.3I/2a}\] \[{{B}_{1}}:{{B}_{2}}=1:3\]


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