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° with \[\overrightarrow{B}.\]The magnetic flux linked with the disc is

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