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A metallic ring of field. If z is the radius l (ring being horizontal is falling under gravity in a region haivng 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 the resistance of the ring and if the ring falls with a velocity `upsilon`, find the energy lost in the resistance If the ring has reached a constant velocity, use the conservation of energy to determine `upsilon` in terms of m, B, `lambda` and acceleration due to gravity g. |
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Answer» Here, `B_(z) = B_(0) (1 + lambda z)` As the metallic ring falls under gravity in this magnetic field, magnetic flux linked with the ring changes Therefore, an emf is induced in the ring. If I is induced current in the ring at any instant, then induced emf ` = IR = (d phi)/(dt) = A xx (dB_(z))/(dt) = (pi l^(2)) B_(0) lambda (dz)/(dt)` `:. I = (pi l^(2) B_(0) lambda upsilon)/(R )`, where `(dz)/(dt) = upsilon` Energy `lost//sec` in the from of heat produced in the ring `= I_(2) R = ((pi l^(2) lambda)^(2) B_(0)^(2) upsilon^(2))/(R )` This comes from rate of change of potential energy of ring `= mg (dz)/(dt) = mg upsilon` `:. ((pi l^(2) lambda)^(2) B_(0)^(2) upsilon^(2))/(R ) = mg upsilon` or `upsilon = (mg R)/((pi l^(2) lambda B_(0) )^(2))` |
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