question_answer

A circular disc of radius 0.2 m is placed in a uniform magnetic field of induction \[\frac{1}{\pi }\left( \frac{Wb}{{{m}^{2}}} \right)\] in such a way that its axis makes an angle of \[{{60}^{o}}\] with \[\overrightarrow{B}\]. The magnetic flux linked with the disc is                                                                                                           [AIPMPT (S) 2008]

A)  0.02 Wb        

B)  0.06 Wb

C)  0.08 Wb        

D)       0.01 Wb

Correct Answer: A

Solution :

            [a] 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, \[\phi =\frac{1}{\pi }\times \pi {{(0.2)}^{2}}\times \cos {{60}^{o}}\] \[={{(0.2)}^{2}}\times \frac{1}{2}\] \[=0.02\,Wb\]