A) \[1\times {{10}^{-3}}\,N\]
B) \[2\times {{10}^{-3}}\,N\]
C) \[4\times {{10}^{-3}}\,N\]
D) \[0.25\times {{10}^{-3}}\,N\]
Correct Answer: C
Solution :
\[F=\frac{{{\mu }_{0}}}{4\pi }\frac{2{{i}_{1}}{{i}_{2}}}{a}={{10}^{-3}}N\] When current in both the wires is doubled, then \[F'=\frac{{{\mu }_{0}}}{4\pi }\frac{2\left( 2{{i}_{1}}\times 2{{i}_{2}} \right)}{a}=4\times {{10}^{-3}}N\]
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