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  • A metallic ring of mass m and radius l (ring being horizontal) is falling under gravity in a region having a magnetic- field. If z is the vertical direction, the z-component of magnetic field is B_{z} = B_{0} (1+ \lambda z).

 A metallic ring of mass m and radius l (ring being horizontal) is falling under gravity in a region having a magnetic- field. If z is the vertical direction, the z-component of magnetic field is B_{z} = B_{0} (1+ \lambda z). If R is the resistance of the ring and if the ring falls with a velocity v, find the energy lost in the resistance. If the ring has reached a constant velocity, use the conservation of energy to determine v in terms of m, B, \lambda and acceleration due to gravity g.

Answers (1)

Here a relation is established between the current induced, velocity of free falling ring and power lost. Let the mass of ring be m and radius l;

\phi _{m}=B_{z}(\pi l^{2})=B_{0}(1+\lambda z)(\pi l)

Applying Faraday's law of EMI, we have emf induced given by

\frac{d\phi }{dt}= rate of change of flux. Also, by Ohm's law

B_{0}(\pi l^{2})\lambda \frac{dz}{dt}=IR

We have

I=\frac{\pi l^{2}B_{0}\lambda }{R}v

Energy lost/second =

I^{2}R=\frac{(\pi l^{2}\lambda )^{2}B_{0}^{2}v^{2}}{R}

Rate of change of

PE=mg\frac{dz}{dt}=mgv [ as kinetic energy is constant for v=constant]

According to the law of conservation of energy

Thus,

mgv=\frac{(\pi l^{2}\lambda B_{0})^{2}v^{2}}{R} or

v=\frac{mgR}{(\pi l^{2}\lambda B_{0})^{2}}

This is the required expression of velocity.

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