A) \[1.44\times {{10}^{-6}}Wb\]
B) \[1.732\times {{10}^{-6}}Wb\]
C) \[3.1\times l{{0}^{-6}}Wb\]
D) \[4.2\times {{10}^{-6}}Wb\]
Correct Answer: B
Solution :
Magnetic flux,\[\phi =\int{\overrightarrow{B}}.\overrightarrow{dA}=BA\cos \theta ,\], where\[\theta \] is angle between normal to the area dA with magnetic field B. Here, \[\theta =({{90}^{o}}-{{30}^{o}})\] \[={{60}^{o}}\] and \[\phi ={{10}^{-4}}\times \pi {{\left[ \frac{21}{2}\times {{10}^{-2}} \right]}^{2}}\times \cos {{60}^{o}}\] \[=1.732\times {{10}^{-6}}Wb\]
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