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A rectangular wire frame of dimensions (0.25 xx 2.0 m) and mass 0.5 kg falls from a height 5m above a region occupied by uniform magnetic field of magnetic induction 1 T. The resistance of the wire frame is 1/8 (Omega). Find time taken by the wire frame when it just starts coming out of the magnetic field. |
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Answer» 0.2s `V_(1) sqrt(2 gh) = sqrt(2(10)(5))= 10m//s` and the time taken is `t, = sqrt((2h)/(g) = sqrt((2(5))/(10))=1s` . (ii) When the frame has partially entered the field, the induced emf produced is `epsilon =BLV` `I=(epsilon)/(R) = (Blv)/(R)` (anticlockwise) Ampere's force, `F=(B^(2)l^(2)v)/(R)` (upward) Putting `m=0.5 kg, B=1 T, l=0.25m, `v=10m//s` R=1//8 Omega` We get, `F=((1)^(2)(0.25)^(2)(10))/(1//8) =5N` Since, `mg=(0.5)(10) = 5N`  Since, `mg=(0.5)(10)=5N` Therefore, using Newton's second LAW, the acceleration of the wire frame while entering the magnetic field is zero, thus, time taken to completely entering into the field is `t_(2)=2/10 =0.2 s` (iii) When the frame has completely entered the field, current become zero and thus, the ampere's force ALSO become zero. The frame accelerates under gravity only. `:. 15 = 10t_(3)+5t_(3)2` or, `t_(3)^(2)+2t_(3)-3 =0` or `t_(3)=1s` The total time taken is `T=t_(1)+t_(2)+t_(3)=1 + 0.2 + 1 = 2.2 s`. |
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