A) \[\begin{align} & 12.6\times {{10}^{-3}}\text{Wb/}{{\text{m}}^{\text{2}}} \\ & 6.3\times {{10}^{-3}}\text{Wb/}{{\text{m}}^{\text{2}}} \\ \end{align}\]
B) \[\begin{align} & 12.6\times {{10}^{-3}}\text{Wb/}{{\text{m}}^{\text{2}}} \\ & 25.1\times {{10}^{-3}}\text{Wb/}{{\text{m}}^{\text{2}}} \\ \end{align}\]
C) \[\begin{align} & 25.1\times {{10}^{-3}}\text{Wb/}{{\text{m}}^{\text{2}}} \\ & 12.6\times {{10}^{-3}}\text{Wb/}{{\text{m}}^{\text{2}}} \\ \end{align}\]
D) \[\begin{align} & 25.1\times {{10}^{-5}}\text{Wb/}{{\text{m}}^{\text{2}}} \\ & 12.6\times {{10}^{-5}}\text{Wb/}{{\text{m}}^{\text{2}}} \\ \end{align}\]
Correct Answer: C
Solution :
Given, \[N=50\text{ }turns/cm=5000\text{ }tums/m\] \[I=4A\] Magnetic field at an internal point \[={{\mu }_{0}}nI\] \[=4\pi \times {{10}^{-7}}\times 5000\times 4\] \[=8\pi \times {{10}^{-3}}\] \[=25.12\times {{10}^{-3}}Wb/{{m}^{2}}\] 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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