A) \[3.6\times {{10}^{-5}}Wb/{{m}^{2}}\]
B) \[2.56\times {{10}^{-4}}Wb/{{m}^{2}}\]
C) \[3.50\times {{10}^{-4}}Wb/{{m}^{2}}\]
D) \[5.80\times {{10}^{-4}}Wb/{{m}^{2}}\]
Correct Answer: B
Solution :
\[B=\frac{{{\mu }_{0}}}{4\pi }\frac{\mu }{{{\left[ {{\left( \frac{L}{2} \right)}^{2}}+{{y}^{2}} \right]}^{3/2}}}\] \[{{B}_{1}}={{10}^{-7}}\times \frac{1.2}{{{\left( {{(0.1)}^{2}}+{{(0.005)}^{2}} \right)}^{3/2}}}\] \[={{10}^{-7}}\frac{(1.2)}{{{10}^{-3}}}=1.2\times {{10}^{-4}}\] \[Wb/{{m}^{2}}\](South to North) \[{{B}_{2}}=1\times {{10}^{-4}}Wb/{{m}^{2}}\] (south to North) \[{{B}_{H}}=3.6\times {{10}^{-5}}Wb/{{m}^{2}}\] (South to North) \[{{B}_{net}}=(0.36+1+1.2)\times {{10}^{-4}}=2.56\times {{10}^{-4}}\] \[Wb/{{m}^{2}}\]You need to login to perform this action.
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