A) \[1.6\times {{10}^{-3}}T\]
B) \[3.2\times {{10}^{-3}}T\]
C) \[4.8\times {{10}^{-3}}T\]
D) \[6.4\times {{10}^{-3}}T\]
Correct Answer: C
Solution :
The straight portions of the wires do not contribute because the point P is along them. The field at P is due to 3/4th of the loop of radius r. Thus, \[B=\frac{3}{4}\left( \frac{{{\mu }_{0}}I}{2r} \right)=\frac{3}{4}\times \frac{4\pi \times {{10}^{-7}}\times 40}{3.14\times {{10}^{-2}}}=4.8\times {{10}^{-3}}T\] Hence, the correction option is [c].
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