A magnet is suspended in the magnetic meridian with an untwisted wire. The upper end of wire is rotated through $$180o$$ to deflect the magnet by $$30o$$ from magnetic meridian. When this magnet is replaced by another magnet, the upper end of wire is rotated through $$270o$$ to deflect the magnet $$30o$$ from magnetic meridian. The ratio of magnetic moments of magnets is

Question

A magnet is suspended in the magnetic meridian with an untwisted wire. The upper end of wire is rotated through $$180o$$ to deflect the magnet by $$30o$$ from magnetic meridian. When this magnet is replaced by another magnet, the upper end of wire is rotated through $$270o$$ to deflect the magnet $$30o$$ from magnetic meridian. The ratio of magnetic moments of magnets is

Infinity Learn · Accepted Answer

Let $$M1$$ and $$M2$$ be the magnetic moments of magnets and BH the horizontal component of earth's field. We have τ$$=MBH$$ $$sin$$θ. if ϕ is the twist of wire, then τ$$=C$$ϕ, C being restoring couple per unit twist of wire⇒ Cϕ$$=$$ MH $$sin$$θHere, ϕ$$1=180o-30o=150o=150$$ x π$$180$$ radϕ$$2=270o-30o=240o=240$$ x π$$180$$ radSo, Cϕ$$1$$ $$=$$ $$M1BH$$ $$sin$$ θ (For$$ deflection θ$$=30o$$ of I $$magnet) Cϕ$$2$$ $$=$$ $$M2BH$$ $$sin$$ θ (For$$ deflection θ$$=30o$$ of II $$magnet)⇒ϕ$$1$$ϕ$$2=M1M2$$⇒$$M1M2=$$ϕ$$1$$ϕ$$2=150$$ x π$$180240$$ x π$$180=1524=58$$⇒$$M1$$ : $$M2=5$$ : $$8$$

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