A) \[\begin{align} & \text{12}\text{.6 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{Wb/}{{\text{m}}^{\text{2}}} \\ & \text{6}\text{.3 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{Wb/}{{\text{m}}^{\text{2}}} \\ \end{align}\]
B) \[\begin{align} & \text{12}\text{.6 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{Wb/}{{\text{m}}^{\text{2}}} \\ & 25.1\times {{10}^{-3}}Wb/{{m}^{2}} \\ \end{align}\]
C) \[\begin{align} & \text{25}\text{.1 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{Wb/}{{\text{m}}^{\text{2}}} \\ & \text{12}\text{.6 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{Wb/}{{\text{m}}^{\text{2}}} \\ \end{align}\]
D) \[\begin{align} & \text{25}\text{.1 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-5}}}\text{Wb/}{{\text{m}}^{\text{2}}} \\ & \text{12}\text{.6 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-5}}}\text{Wb/}{{\text{m}}^{\text{2}}} \\ \end{align}\]
Correct Answer: C
Solution :
Given,\[N=50\,\,turns/cm=5000\,\,turns/cm\] \[I=4A\] Magnetic field at an internal point\[={{\mu }_{0}}nI\] \[=4\pi \times {{10}^{-7}}\times 5000\times 4\] \[=25.12\times {{10}^{-3}}Wb/{{m}^{2}}\] Magnetic field at one end\[=\frac{{{\mu }_{0}}nI}{2}\] \[=\frac{25.12\times {{10}^{-3}}}{2}\] \[=12.56\times {{10}^{-3}}Wb/{{m}^{2}}\]
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