At a certain location in the northern hemisphere, the earth's magnetic field has magnitude of `42 mu T` and points downwards at `53^(@)` to the vertical. Calculate the flux through a horizontal surface of area `2.5 m^(2).[sin 53^(@) = 0.8]`

Explanation

`phi_(B) = BA cos theta`
`= 42 xx 10^(-6) xx 2.5 xx cos 53^(@) = 63 mu Wb`