A) \[6.67\times {{10}^{3}}Wb\]
B) \[6.67\times {{10}^{4}}Wb\]
C) \[3.67\times {{10}^{4}}Wb\]
D) \[3.67\times {{10}^{3}}Wb\]
Correct Answer: B
Solution :
\[{{\phi }_{q}}=\frac{{{\mu }_{0}}{{i}_{1}}{{R}^{2}}}{2{{({{R}^{2}}+{{x}^{2}})}^{\frac{3}{2}}}}\times \pi {{r}^{2}}={{10}^{-3}}\] \[{{\phi }_{p}}=\frac{{{\mu }_{0}}{{i}_{2}}{{r}^{2}}}{2{{({{r}^{2}}+{{x}^{2}})}^{\frac{3}{2}}}}\times \pi {{R}^{2}}\] \[\frac{{{\phi }_{p}}}{{{\phi }_{Q}}}=\frac{{{i}_{2}}}{{{i}_{1}}}.\frac{{{({{R}^{2}}+{{x}^{2}})}^{\frac{3}{2}}}}{{{({{r}^{2}}+{{x}^{2}})}^{\frac{3}{2}}}}=\frac{{{\phi }_{p}}}{{{10}^{-3}}}\] \[\frac{2}{3}=\frac{{{\phi }_{p}}}{{{10}^{-3}}}\] \[{{\phi }_{p}}=6.67\times {{10}^{-4}}\].
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