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O B ST 252 3. SWITCH S",s CLOSEN AT t=0.Tite BAR HAS AMASS, m3 2RggAND IS INITIALLY AT ResT. Fin : a) Vronia b) E MAX c) i IND MAX 2) L3 MAX IND MAX ) P IND MAX h) TIMENGEDEU TU REACH VTERM 2. O DISTANCE TEAYELED UPTO THIS PUINT INTIME = ) ACCELERRATION AT THIS TIME =

O B ST 252 3. SWITCH S",s CLOSEN AT t=0.Tite BAR HAS AMASS, m3 2RggAND IS INITIALLY AT ResT. Fin : a) Vronia b) E MAX c) i IND MAX 2) L3 MAX IND MAX ) P IND MAX h) TIMENGEDEU TU REACH VTERM 2. O DISTANCE TEAYELED UPTO THIS PUINT INTIME = ) ACCELERRATION AT THIS TIME =

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O B ST 252 3. SWITCH S",s CLOSEN AT t=0.Tite BAR HAS AMASS, m3 2RggAND IS INITIALLY AT ResT. Fin : a) Vronia b) E MAX c) i IND MAX 2) L3 MAX IND MAX ) P IND MAX h) TIMENGEDEU TU REACH VTERM 2. O DISTANCE TEAYELED UPTO THIS PUINT INTIME = ) ACCELERRATION AT THIS TIME =
Transcribed Image Text:O B ST 252 3. SWITCH S",s CLOSEN AT t=0.Tite BAR HAS AMASS, m3 2RggAND IS INITIALLY AT ResT. Fin : a) Vronia b) E MAX c) i IND MAX 2) L3 MAX IND MAX ) P IND MAX h) TIMENGEDEU TU REACH VTERM 2. O DISTANCE TEAYELED UPTO THIS PUINT INTIME = ) ACCELERRATION AT THIS TIME =
Expert Solution
Step 1

a)

 terminal velocity, Vterm =2 Bvl5 here Bvl = 1000Vterm = 400m/s

b)

maximum induced emf, ∈ind(max) = 1000 V

c)

maximum induced current, Iind(max) = ∈indR = 10002 =500 A

 

Step 2

d)

dϕdtmax =maximum induced emf=  1000 V

e)

maximum force induced,Find(max) = Iind(max) × LB= 500 ×12 ×5 = 1250 N

f)

maximum power induced, Pind(max) = Iind2 × R = 5002 ×2 = 5000000 W

g)

a0 = 10002 ×12 ×52 = 625 m/s2

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