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 \[\frac{{{\mu }_{0}}{{I}_{1}}I}{2\pi x}+\frac{{{\mu }_{0}}{{I}_{1}}I}{2\pi (d-x)}=0\] \[\frac{{{I}_{1}}}{x}=\frac{{{I}_{2}}}{x-d}\] \[\Rightarrow x=\frac{{{I}_{1}}d}{{{I}_{1}}-{{I}_{2}}}\]You need to login to perform this action.
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