A) \[\frac{1}{2}\]
B) 2
C) \[\frac{2}{3}\]
D) \[\frac{3}{2}\]
E) \[\frac{1}{5}\]
Correct Answer: D
Solution :
When the currents are in the same direction then magnetic field \[B=\frac{{{\mu }_{0}}}{4\pi }\times \frac{1}{d}[{{i}_{}}_{1}-{{i}_{2}}]\] \[6\times {{10}^{-6}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2}{d}[{{i}_{1}}-{{i}_{2}}]\] ...(i) When the currents are in the reversed direction then magnetic field \[{{B}_{2}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2}{d}[{{i}_{1}}-(-{{i}_{2}})]\] Or \[{{B}_{2}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2}{d}[{{i}_{1}}+{{i}_{2}}]\] or \[3\times {{10}^{-5}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2}{d}[{{i}_{1}}+{{i}_{2}}]\] ...(ii) Dividing Eq. (i) by Eq. (ii) \[\frac{{{i}_{1}}-{{i}_{2}}}{{{i}_{1}}+{{i}_{2}}}=\frac{6\times {{10}^{-6}}}{3\times {{10}^{-5}}}\] Or \[\frac{{{i}_{1}}-{{i}_{2}}}{{{i}_{1}}+{{i}_{2}}}=\frac{2}{10}\] Or \[5{{i}_{1}}-5{{i}_{2}}={{i}_{1}}+{{i}_{2}}\] Or \[4{{i}_{1}}=6{{i}_{2}}\] Or \[\frac{{{i}_{1}}}{{{i}_{2}}}=\frac{6}{4}\] Or \[\frac{{{i}_{1}}}{{{i}_{2}}}=\frac{3}{2}\]
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