A bar magnet of magnetic moment `M_1` is suspended by a wire in a magnetic field. The tip of the wire is rotated through ` 180^(@)` , then the magnet rotated through ` 45^(@)` . Under similar conditions the magnet of magnetic moment `M_2` is rotated through ` 30^@` . Then find the ratio of `M_2 & M_2`.

Explanation

We know that torque `, tau = MB sin theta `
But ` tau = C phi` .
`:. C phi = MB sin theta `
For first magnet ,
` C xx 180 = M_1B sin 45^(@)` ....... (1)
For second magnet ,
` C xx 180 = M_2B sin30^(@)` ....... (2)
From eqs. (1) and (2)
`M_1 B sin 45^(@) = M_2 B sin 30^(@) implies (M_1)/(M_2) = (sin 30^@)/(sin 45^(@))`
`implies (M_1)/(M_2) = (1//2)/(1 // sqrt(2)) = (sqrt(2))/(2) = (1)/(sqrt(2))`
`:. M _1 : M_2 = 1: sqrt(2)`