JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer Two wires A & B are carrying currents \[{{I}_{1}}\And {{I}_{2}}\]as shown in the figure. The separation between them is d. A third wire C carrying a current I is to be kept parallel to them at a distance x from A such that the net force acting on it is zero. The possible values of x are: [JEE Main 10-4-2019 Morning]

    A) \[x=\left( \frac{{{I}_{1}}}{{{I}_{1}}-{{I}_{2}}} \right)d\]and\[x=\frac{{{I}_{2}}}{({{I}_{1}}+{{I}_{2}})}d\]

    B) \[x=\pm \frac{{{I}_{1}}d}{({{I}_{1}}-{{I}_{2}})}\]

    C) \[x=\left( \frac{{{I}_{1}}}{{{I}_{1}}+{{I}_{2}}} \right)d\]and\[x=\frac{{{I}_{2}}}{({{I}_{1}}-{{I}_{2}})}d\]

    D) \[x=\left( \frac{{{I}_{2}}}{{{I}_{1}}+{{I}_{2}}} \right)d\]and\[x=\left( \frac{{{I}_{2}}}{{{I}_{1}}-{{I}_{2}}} \right)d\]

    Correct Answer: B

    Solution :

    Net force on wire carrying current I per unit length is \[\frac{{{\mu }_{0}}{{I}_{1}}I}{2\pi x}+\frac{{{\mu }_{0}}{{I}_{1}}I}{2\pi (d-x)}=0\] \[\frac{{{I}_{1}}}{x}=\frac{{{I}_{2}}}{x-d}\] \[\Rightarrow x=\frac{{{I}_{1}}d}{{{I}_{1}}-{{I}_{2}}}\]


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