At a place the earth's horizontal component of magnetic field is \[0.36\times {{10}^{-4}}weber/{{m}^{2}}\]. If the angle of dip at that place is 60o, then the vertical component of earth's field at that place in weber/m2 will be approximately [MP PMT 1985]

• A.            \[0.12\times {{10}^{-4}}\]      
• B.            \[0.24\times {{10}^{-4}}\]
• C.            \[0.40\times {{10}^{-4}}\]      
• D.            \[0.62\times {{10}^{-4}}\]