A) \[\frac{{{\mu }_{0}}{{I}_{1}}{{I}_{2}}}{4\pi }\,\frac{(x_{2}^{2}+y_{2}^{2})}{(x_{1}^{2}+y_{1}^{2})}\]
B) \[\frac{{{\mu }_{0}}{{I}_{1}}{{I}_{2}}}{4\pi }\,In\,\frac{(x_{2}^{2}+y_{2}^{2})}{(x_{1}^{2}+y_{1}^{2})}\]
C) \[\frac{{{\mu }_{0}}{{I}_{1}}{{I}_{2}}}{4\pi }\,In\,\frac{({{x}_{2}}+{{y}_{2}})}{({{x}_{1}}+{{y}_{1}})}\]
D) \[\frac{{{\mu }_{0}}{{I}_{1}}{{I}_{2}}}{2\pi }\,In\,\frac{(x_{2}^{2}+y_{2}^{2})}{(x_{1}^{2}+y_{1}^{2})}\]
Correct Answer: B
Solution :
[b] \[dF=\frac{{{\mu }_{0}}{{I}_{1}}{{I}_{2}}dr}{2\pi r}\Rightarrow F=\int\limits_{{{r}_{1}}}^{{{r}_{2}}}{\frac{{{\mu }_{0}}{{I}_{1}}{{I}_{2}}}{2\pi r}}dr=\frac{{{\mu }_{0}}{{I}_{1}}{{I}_{2}}}{2\pi }[\ell nr]_{{{r}_{1}}}^{{{r}_{2}}}\] \[=\frac{{{\mu }_{0}}{{I}_{1}}{{I}_{2}}}{2\pi }\left[ \ell n{{r}_{2}}-\ell n{{r}_{1}} \right]=\frac{{{\mu }_{0}}{{I}_{1}}{{I}_{2}}}{4\pi }\ell n\left( \frac{x_{2}^{2}+y_{2}^{2}}{x_{1}^{2}+y_{1}^{2}} \right)\]You need to login to perform this action.
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