A) \[\text{1 }\times \text{ 1}{{0}^{-\text{3}}}\text{ N}\]
B) \[\text{2 }\times \text{ 1}{{0}^{\text{-3}}}\text{ N}\]
C) \[~\text{4 }\times \text{ 1}{{0}^{-\text{3}}}\text{ N}\]
D) \[0.\text{25 }\times \text{ 1}{{0}^{-\text{3}}}\text{ N}\]
Correct Answer: C
Solution :
\[F=\frac{{{\mu }_{0}}}{4\pi }\cdot l\frac{2{{i}_{1}}{{i}_{2}}}{d}\] \[\therefore \] \[1\times {{10}^{-3}}=\frac{{{\mu }_{0}}}{4\pi }l\frac{2\times 10\times 10}{d}\] ? (i) and \[F=\frac{{{\mu }_{0}}}{4\pi }l\frac{2\times 20\times 20}{d}\] ? (ii) From equations (i) and (ii), \[\frac{1\times {{10}^{-3}}}{F}=\frac{\frac{{{\mu }_{0}}}{4\pi }l\frac{2\times 10\times 10}{d}}{\frac{\mu }{4\pi }\frac{2\times 20\times 20}{d}}=\frac{10\times 10}{20\times 20}\] \[\therefore \] \[\mathbf{F=4\times 1}{{\mathbf{0}}^{\mathbf{-3}}}\mathbf{N}\].
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