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`        Consider a non-conducting ring of radius r and mass m which has a total charge q distributed uniformly on it. The ring is rotated about its axis with an angular speed ω. (a) Find the equivalent electric current in the ring. (b) Find the magnetic moment μ of the ring. (c) Show that μ = q/2m l where l is the angular momentum of the ring about its axis of rotation.`
3 years ago

396 Points
```										Sol. (a) i = q/t = q/(2π / ω) = qω/2π
(b) μ = n ia = i A [∴ n = 1] = qωπ r^2/2π = qωr^2/2
(c) μ = qωr^2/2, L = Iω = mr^2 ω, μ/L = qωr^2/2mr^2ω = q/2m ⇒ μ = (q/2m)L

```
3 years ago
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