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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_O (I + lambdaz). 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, lambdaand acceleration due to gravity g. |
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Answer» Solution :Magnetic flux linked with ring, `phi=AB costheta (because theta =0)` `theta =Bzpil^2` `phi=B_0(1+lambdaz)pil^2` Now according to Faraday.s law induced emf, `epsilon=(dphi)/(DT)=d/(dt)[B_0 (1+lambdaz)pil^2]` `epsilon=B_0 pil^2 lambda (dz)/(dt)` `epsilon=B_0 pil^2 lambdav` Induced current `I=S/R` `I=(B_0pil^2lambdav)/R` Heat produced PER unit time, `H=I^2R=((Bpil^2lambdav)/R)^2R` `H=(B_0^2pi^2l^4lambda^2v^2)/R` This heat per unit time is produced at the cost of rate of change of potential energy. Rate of change of P.E. = Heat produced per unit time (`because`Here that K.E. remains constant) `d/(dt)(mgz)=H` `mg(dz)/(dt)=(B_0^2pi^2l^4lambda^2v^2)/R` `THEREFORE mgv=(B_0^2pi^2l^4lambda^2v^2)/R` `therefore v=(mgR)/(B_0^2pi^2l^4lambda^2)` which is required equation of terminal velocity . |
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