A circular disc of radius `0.2 m` is placed in a uniform magnetic fied of induction `(1)/(pi) ((Wb)/(m^(2)))`
in such a way that its axis makes an angle of `60^(@)` with The magnetic flux linked with the disc is

Options

A. 0.02 Wb
B. 0.06 Wb
C. 0.08 Wb
D. 0.01 Wb

Answer

A

Explanation

The magnetic flux `phi` passing through 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^(@),B=(1)/(pi)Wbm^(-2),A=pi(0.2)^(2)`
`therefore phi=(1)/(pi)xxpi(0.2)^(2)xxcos60^(@)=0.02Wb`