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Electromagnetic Induction — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsElectromagnetic Induction5 Marks
Question
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 $\text{B}_\text{z}=\text{B}_0(1+\lambda\text{z})$. 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 $\text{m},\text{B},\lambda$ and acceleration due to gravity g.
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Answer

In this problem a relation is established between induced current, power lost and velocity acquired by freely falling ring.
The magnetic flux linked with the metallic ring of mass m and radius l ring being horizontal falling under gravity in a region having a magnetic field whose z-component of magnetic field is $\text{B}_\text{z}=\text{B}_0(1+\lambda\text{z})$ is,
$\phi_\text{m}=\text{B}_\text{z}(\pi\text{l}^2)=\text{B}_0(1+\lambda\text{z})(\pi\text{l})$
Applying Faraday's law of EML, we have emf induced given by $\frac{\text{d}\phi}{\text{dt}}$ = rate of change of flux. Also, by Ohm's law
$\text{B}_0(\pi\text{l}^2)\lambda\frac{\text{dz}}{\text{dt}}=\text{IR}$
We have $\text{I}=\frac{\pi\text{l}^2\text{B}_0\lambda}{\text{R}}\text{v}$
Energy lost/second $=\text{I}^2\text{R}=\frac{(\pi\text{l}^2\lambda)^2\text{B}_0^2\text{v}^2}{\text{R}}$
Rate of change of $\text{PE}=\text{mg}\frac{\text{dz}}{\text{dt}}=\text{mgv}$ [as kinetic energy is constant for v = constant]
According to the law of conservation of energy
Thus, $\text{mgv}=\frac{(\pi\text{l}^2\lambda\text{B}_0)^2\text{v}^2}{\text{R}}\text{ or v}=\frac{\text{mgR}}{(\pi\text{l}^2\lambda\text{B}_0)^2}$
This is the required exspression of velocity.

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