A circular loop wire with current \[{{i}_{1}}=({{i}_{0}}/\pi )\] and a V shaped wire with current;, are arranged in a plane as shown in diagram. If magnetic field at O is zero, value of \[{{\operatorname{i}}_{2}}\] is: |
A) \[\frac{13a{{i}_{0}}}{35r}\]
B) \[\frac{12a{{i}_{0}}}{17r}\]
C) \[\frac{a{{i}_{0}}}{r}\]
D) \[\frac{12a{{i}_{0}}}{35r}\]
Correct Answer: D
Solution :
here, \[a=r'\left( \cot 37{}^\circ +\cot 53{}^\circ \right)\] |
\[={{r}^{'}}\left( \frac{4}{3}+\frac{3}{4} \right)\] |
\[\therefore {{r}^{'}}=\frac{12a}{25}\] |
Magnetic field due to-V-shaped conducter |
\[{{B}_{1}}=\left[ \frac{{{\mu }_{0}}}{4\pi }.\frac{{{i}_{2}}}{\frac{12a}{25}}\left( \sin 53{}^\circ +\sin 37{}^\circ \right) \right]\]\[=2\left[ \frac{{{\mu }_{0}}}{4\pi }.\frac{{{i}_{2}}}{\frac{12a}{25}}\times \frac{7}{5} \right]\] or \[70\left( \frac{{{\mu }_{0}}}{4\pi } \right)\frac{{{i}_{2}}}{12a}\] |
According to given condition |
\[{{B}_{1}}={{B}_{2}}\] |
\[70\left( \frac{{{\mu }_{0}}}{4\pi } \right)\frac{{{i}_{2}}}{12a}=\frac{{{\mu }_{0}}\left( \frac{{{i}_{0}}}{\pi } \right)}{2r}\therefore {{i}_{2}}=\frac{12a{{i}_{0}}}{35r}.\] |
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