Charge Q is uniformly distributed on a thin insulating ring of mass m which is initially at rest. To what angular velocity will the ring be accelerated when a magnetic field B, perpendicular to the plane of the ring, is switched on?
Text Solution
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Sol. The changing magnetic field induces an electric field in the ring. Let us imagine the ring divided into small sections each of length Δ s and denote the tangential component of the induced electric field by E t (in the general case E t can vary from point to point). The charge on a small section of the ring is
Δ Q = Q
,
where r is the radius of the ring. The force exerted on it is
Δ F t = Δ Q E t
and the resultant torque is
Δτ = r Δ F t .
The total torque experienced by the ring is thus
τ =
=
Q
E t =
.
Identifying the expression Σ E t Δ s as the induced electromotive force along the ring, which is directly proportional to the rate of change in the magnetic flux, we have
= –
= – π r 2
.
As a result of the torque, the ring, which has a moment of inertia I = mr 2 , starts to spin with angular acceleration α . During a time interval Δ t its angular velocity changes by
Δω = αΔ t =
Δ t =
Δ t = –
Δ B.
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