A) 0.02 Wb
B) 0.06 Wb
C) 0.08 Wb
D) 0.01 Wb
Correct Answer: A
Solution :
The magnetic flux \[\phi \] passing through a plane surface of area A placed in a uniform magnetic field B is given by\[\phi =BA\cos \theta \]where \[\theta \] is the angle between the direction of B and the normal to the plane. Here,\[\theta ={{60}^{o}},B=\frac{1}{\pi }Wb/{{m}^{2}},A=\pi {{(0.2)}^{2}}\] Therefore, \[\phi =\frac{1}{\pi }\times \pi {{(0.2)}^{2}}\times \cos {{60}^{o}}\] \[={{(0.2)}^{2}}\times \frac{1}{2}\] \[=0.02Wb\]You need to login to perform this action.
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