A) \[16\times {{10}^{-5}}A\]
B) \[16A\]
C) \[4A\]
D) \[1A\]
Correct Answer: C
Solution :
(c) \[4A\] Here \[{{l}_{1}}={{l}_{2}}=l,\,F=16\times {{10}^{-5}}N/m,\,d=2\,cm\] \[F=\frac{{{\mu }_{0}}2{{l}_{1}}{{l}_{2}}}{4\pi d}=\frac{{{\mu }_{0}}}{4\mu }\frac{2{{l}^{2}}}{d}\] \[16\times {{10}^{-5}}={{10}^{-7}}\times \frac{2\times {{l}^{2}}}{2\times {{10}^{-2}}}\] \[\Rightarrow {{l}^{2}}=\frac{16\times {{10}^{-5}}}{{{10}^{-5}}}=16\Rightarrow l=4\,A\].You need to login to perform this action.
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