A) 0.036 A
B) 0.9 A
C) 0.45 A
D) 4.5 A
Correct Answer: D
Solution :
Let due to be a straight conductor carrying a current of i ampere, magnetic field at point is \[B=\frac{{{\mu }_{0}}}{4\pi }\frac{i}{R}(sino{{|}_{1}}+sino{{|}_{2}})\] For a conductor of infinite length, we have \[\text{o}{{\text{ }\!\!|\!\!\text{ }}_{1}}=\text{o}{{\text{ }\!\!|\!\!\text{ }}_{2}}={{90}^{o}}\] \[\therefore \] \[B=\frac{{{\mu }_{0}}}{4\pi }\frac{2i}{R}\] Given, \[R=5\times {{10}^{-2}}m\] \[\therefore \] \[B={{10}^{-7}}\left( \frac{2i}{5\times {{10}^{-2}}} \right)\] \[=(4i)\times {{10}^{-2}}\]gauss At neutral point, \[4i\times {{10}^{-2}}=0.18\] \[\Rightarrow \] \[i=4.5\,A\]
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