A) 0.29 A
B) 1.2 A
C) 3.6 ´ 10?5 A
D) 0.14 A
Correct Answer: A
Solution :
\[i=\frac{2r{{B}_{H}}}{{{\mu }_{0}}N}\tan \theta \] \[\Rightarrow i=\frac{2\times 15\times {{10}^{-2}}\times 3\times {{10}^{-5}}}{4\pi \times {{10}^{-7}}\times 25}\times \tan {{45}^{o}}\]\[\Rightarrow i=0.29\ A\]
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