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A bar magnet of magnetic moment M_1 is suspended by a wirein a magnetic field. The tip of the wire is rotated through180^(@) , then the magnet rotated through45^(@) . Under similar conditions the magnet of magnetic moment M_2 is rotated through30^@ . Then find the ratio of M_2& M_2. |
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Answer» Solution :We knowthat 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)` |
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