Question

A magnetic field B is confined to a region r ≤ a and points out of the paper (the z-axis), r = 0 being the centre of the circular region. A charged ring (charge = Q) of radius b, b > a and mass m lies in the x-y plane with its centre at the origin. The ring is free to rotate and is at rest. The magnetic field is brought to zero in time ∆t. Find the angular velocity ω of the ring after the field vanishes.

Answer 1

According to the law of EMI, when magnetic field changes in the circuit, then magnetic flux linked with the circuit also changes and this changing magnetic flux leads to an induced emf in the circuit.

Here, magnetic field decreases which causes induced emf and hence, electric field around the ring. The torque experienced by the ring produces change in angular momentum.

As the magnetic field is brought to zero in time At, the magnetic flux linked with the ring also reduces from maximum to zero. This, in turn, induces an emf in ring as discussed above. The induces emf causes the induced electric field E around the ring.

By the relation between electric field and potential we get,

The induced emf = Electric field E × (2πb) (Because V = E × d)  ......(i)

By Faraday's law of EMI

`|ε| = (dphi)/(dt) = A (dB)/(dt)`

`|ε| = (Bpia^2)/(Δt) S`  ......(ii)

From equation (i) and (ii), we have

`2pibE = ε = (Bpia^2)/(Δt)`

As we know the electric force experienced by the charged ring, `F_e = QE`. This force try to rotate the coil, and the torque is given by

Torque = b × Force

τ = `QEb = Q[(Bpia^2)/(2piΔt)]b`

⇒ τ = `Q (Ba^2)/(2Δt)`

If ΔL is the change in angular momentum,

ΔL = Torque × Δt = `Q (Ba^2)/2`

Since, initial angular momentum = 0

And  Torque × Δt  = Change in angular momentum

Final angular momentum = `mb^2ω = (QBa^2)/2`

Where, mb² = I (moment of inertia of ring)

`ω = (QBa^2)/(2mb^2)`

On rearranging the terms, we have the required expression of angular speed.