Ncert Exercises & Solutions

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 the magnetic field is $B_{z} = B_{0} (1 + λ z)$ . If R is the resistance of the ring and if the ring has reached a constant velocity, use the conservation of energy to determine v in terms of m, B, $λ$ and acceleration due to gravity g.

Hint: The current is induced due to the change in magnetic flux.

Step 1: The magnetic flux linked with the metallic ring of mass m and radius l falling under gravity in a region having a magnetic field whose z-component of the magnetic field is

$B_{z} = B_{0} (1 + λ z)$ is;

$ϕ = B_{z} ({πl}^{2}) = B_{0} (1 + λz) ({πl}^{2})$

Step 2: Applying Faraday's law of EMl, we have emf induced given by;

$\frac{dϕ}{dt}$ =rate of change of flux=

$B_{0} ({πl}^{2}) λ \frac{dz}{dt}$

Also, by Ohm's law;

$B_{0} ({πl}^{2}) λ \frac{dz}{dt} = IR$

On rearranging the terms, we have;

$I = \frac{{πl}^{2} B_{0} λ}{R} v$

Step 3: Energy lost/second=

$I^{2} R = \frac{{({πl}^{2} λ)}^{2} {B_{0}}^{2} v^{2}}{R}$

This must come from the rate of change in PE=mg

$\frac{dz}{dt}$ =mgv

[as kinetic energy is constant for v=constant]

Thus,

$mgv = \frac{{({πl}^{2} {λB}_{0})}^{2} v^{2}}{R} or v = \frac{mgR}{{({πl}^{2} {λB}_{0})}^{2}}$

This is the required expression of velocity.

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