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JEE Main & Advanced JEE Main Paper (Held On 12 April 2014)

  • 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:   [JEE Main Online Paper ( Held On 12 Apirl  2014 )

    A) 1 : 1                                       

    B) 1 : 3

    C) 1 : 9                                       

    D) 9 : 1

    Correct Answer: B

    Solution :

                    For loop 0nI\[B=\frac{{{\mu }_{0}}nI}{2a}\]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 \[{{B}_{2}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{2\times 3\times \left( I/3 \right)\times \pi {{\left( a/3 \right)}^{2}}}{{{\left( a/3 \right)}^{3}}}\] \[=\frac{{{\mu }_{0}}.3I}{2a}\] \[\therefore \]\[\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{{{\mu }_{0}}I/2a}{{{\mu }_{0}}.3I/2a}\]\[{{B}_{1}}:{{B}_{2}}=1:3\]


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