The work done in turning a magnet of magnetic moment 'M' by an angle of `90^(@)` from the meridian is 'n' times the corresponding work done to turn it through an angle of `60^(@)`, where 'n' is given by

Options

A. `1//2`
B. 2
C. `1//4`
D. 1

Answer

b

Explanation

(b) `W_(1) = - MB(cos 90^(@)-cos 0^(@))=MB`
`W_(2) = - MB (cos 60^(@)-cos 0^(@))`
`= - MB((1)/(2)-1)=(10/(2)MB=(1)/(2)W_(1))`
As `W_(1)=nW_(2), therefore n=2`