Ncert Exercises & Solutions

A magnetic field B is confined to a region r $\leq$ a and points out of the paper (the z-axis), r=0 being the centre of the circular region. 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 field vanishes.

Hint: The change in magnetic flux results in the rotation of the ring.

Since the magnetic field is brought to zero in time

∆

t, the magnetic flux linked with the ring also reduces from maximum to zero. This, in turn, induces an emf in-ring by the phenomenon of EMI. The induced emf causes the electric field E generation around the ring.

Step 1 : The inuced emf = electric field (E) \times (2 πb) (Because V = E \times d) . . . (i)

By Faraday' s law of EMI;

The induced emf = rate of change of magnetic flux

= rate of change of magnetic field \times area

= \frac{{Bπa}^{2}}{∆ t} . . . (ii)

From Eqs . (i) and (ii), we have;

2 πbE = emf = \frac{{Bπa}^{2}}{∆ t}

Step 2 : Since, the charged ring experienced an electric force = QE

This force tries to rotate the coil, and the torque is given by;

Torque = b \times Force

= QEb = Q [\frac{{Bπa}^{2}}{2 πb ∆ t}] b

= Q \frac{{Ba}^{2}}{2 ∆ t}

Step 3 : If ∆ L is the change in angular momentum .

∆ L = Torque \times ∆ t = Q \frac{{Ba}^{2}}{2}

Since, initial angular momentum = 0

Now, since, Torque \times ∆ t = Change in angular momentum

Final angular momentum = {mb}^{2} ω = \frac{{QBa}^{2}}{2}

ω = \frac{{QBa}^{2}}{2 {mb}^{2}}

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

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