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 B_z = B_o (1 + λz). 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, λ and acceleration due to gravity g.

Answer 1

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 B_z = B₀(1 + λz) is `phi = vecB_z.vecA = B_o (1 + λz).pil^2`

The angle between `vecB` and `vecA` is 0°

`ε = d/(dt) [B_o (1 + λz)]pil^2`

`IR = (B_opil^2)[0 + λ (dz)/(dt)]`

`I = (B_opiλl^2)/R (dz)/(dt) = (B_opiλl^2)/R v`

Energy lost = `I^2R = (B_o^2pi^2λ^2l^4)/R^2 v^2R`

Energy lost = `(B_o^2pi^2λ^2l^4v^2)/R`

The energy must come from decrease in P.E = `mg  (dz)/(dt) = mgv`

∴ `mgv = (B_o^2pi^2λ^2v^2l^4)/R`

`v = (mgR)/(B_o^2pi^2λ^2l^4)` or `(mgR)/((pil^2λB_o)^2)`

It is the required relation.