A) 0.02 Wb
B) 0.06 Wb
C) 0.08 Wb
D) 0.01 Wb
Correct Answer: A
Solution :
(a) 0.02 Wb Here, \[\theta =60{}^\circ ,\text{ }B=\frac{1}{\pi }Wb{{m}^{-2}}\text{, }A=\pi {{\left( 0.2\text{ } \right)}^{2}}\] \[\therefore \phi =BA\,\cos \theta \Rightarrow \,\phi =\frac{1}{\pi }\times \pi {{(0.2)}^{2}}\times \cos \,60{}^\circ \] \[={{(0.2)}^{2}}\times \frac{1}{2}=0.02\,Wb\]You need to login to perform this action.
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