A) 2.5 ´ 10?7 weber
B) 6.31 ´ 10?6 weber
C) 5.2 ´ 10?5 weber
D) 4.1 ´ 10?5 weber
Correct Answer: B
Solution :
\[\varphi ={{\mu }_{0}}niA=4\pi \times {{10}^{-7}}\times \frac{3000}{1.5}\times 2\times \pi {{\left( 2\times {{10}^{-2}} \right)}^{2}}\] \[=6.31\times {{10}^{-6}}Wb\]
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