Question

Consider a non-conducting ring of radius r and mass m that 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=q2 m l, where l is the angular momentum of the ring about its axis of rotation.

Answer 1

Explanation:

Provided:

Ring’s radius = r

Ring’s mass = m

Ring’s total charge = q

(a) Angular speed,

ω = 2πT ⇒T = 2πω

Ring’s current,

i = qω2π

(b) The ring with area A and current i, the magnetic moment,

µ = niA = ia  [n = 1]

= qωr22    …(i)

(c) Angular momentum, l = Iω,

where I is ring’s moment of inertia about its rotation axis.

I = mr2So,

⇒ωr2 = lm

When the value in equation (i) is substituted, we obtain:

μ = ql2m