A) \[5\mu T\]
B) \[15\mu T\]
C) \[74\mu T\]
D) \[128\mu T\]
Correct Answer: C
Solution :
[c] Net current is \[\left( 20-6+12-7+18 \right)A\], i.e, 37A \[r=\frac{10}{100}m=\frac{1}{10}m\] \[B=\frac{{{\mu }_{0}}I}{2\pi r}=\frac{4\pi \times {{10}^{-7}}\times 37\times 10}{2\pi \times 1}=74\times {{10}^{-6}}T\] \[74\mu T.\]
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