A) \[15.7\times {{10}^{-7}}Wb/{{m}^{2}}\]
B) \[22.2\times {{10}^{-7}}Wb/{{m}^{2}}\]
C) \[12.6\times {{10}^{-7}}Wb/{{m}^{2}}\]
D) \[18.9\times {{10}^{-7}}Wb/{{m}^{2}}\]
Correct Answer: B
Solution :
The magnetic field at centre of coil P becomes \[{{B}_{1}}=\frac{{{\mu }_{0}}{{i}_{1}}{{N}_{1}}}{2r}=\frac{{{\mu }_{0}}\times 1\times 10}{2\times 4}=\frac{4\pi \times {{10}^{-7}}\times 10}{8}\] \[=5\pi \times {{10}^{-7}}\] Also, \[{{B}_{2}}={{B}_{1}}=5\pi \times {{10}^{-7}}\] \[\therefore \] \[B=\sqrt{B_{1}^{2}+B_{2}^{2}}=5\sqrt{2}\pi \times {{10}^{-7}}Wb/{{m}^{2}}\]
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