A) \[4\times {{10}^{-5}}Wb/{{m}^{2}}\]
B) Zero
C) \[2\times {{10}^{-5}}Wb/{{m}^{2}}\]
D) \[8\times {{10}^{-5}}Wb/{{m}^{2}}\]
Correct Answer: A
Solution :
[a] \[l=4.5\times {{10}^{-2}}m\] |
\[\tan 60{}^\circ =\sqrt{3}=\frac{l}{2d}\] |
\[\Rightarrow d=\frac{l}{2\sqrt{3}}=\left( \frac{4.5\times {{10}^{-2}}}{2\sqrt{3}} \right)m\] |
\[B=\frac{{{\mu }_{o}}i}{4\pi d}(\cos {{\theta }_{1}}+\cos {{\theta }_{2}})\] |
\[=\frac{2{{\mu }_{0}}i}{4\pi d}\left( \frac{\sqrt{3}}{2} \right)\] |
\[=\frac{{{\mu }_{o}}i}{2\pi }\left( \frac{\sqrt{3}}{2} \right)\frac{2\sqrt{3}}{(4.5\times {{10}^{-2}})}\] |
On solving we will get option A as answer |
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