A bar magnet of magnetic moment `M_(1)` is suspended by a wire in a magnetic field. The upper end of the wire is rotated through `180^(@)`, then the magnet rotated through `45^(@)`. Under similar conditions another magnetic of magnetic moment `M_(2)` is rotated through `30^(0)`. Then find the ratio of `M_(1)&M_(2)`.

A bar magnet of magnetic moment `M_(1)` is suspended by a wire in a ma

1.	A bar magnet of magnetic moment `M_(1)` is suspended by a wire in a magnetic field. The upper end of the wire is rotated through `180^(@)`, then the magnet rotated through `45^(@)`. Under similar conditions another magnetic of magnetic moment `M_(2)` is rotated through `30^(0)`. Then find the ratio of `M_(1)&M_(2)`.
Answer» `C(alpha-theta)=MB sin theta` For first magnet, `C(180-45)=M_(1)B sin 45^(@)` For second magnet, `C(180-30)=M_(2)B sin 30^(@)` …….`(2)` Diving equation `(1)` by equation `(2)` `(135)/(150)=(M_(1))/(M_(2))xxsqrt2=rArr(M_(1))/(M_(2))=(9)/(10sqrt2)`

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