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Q.

A particle of charge per unit mass α is released from origin with velocity V→=v0i^ in a magnetic field B→=−B0k^  for  x≤32v0B0α  and B→=0  for x>32v0B0α. Then x–coordinate of the particle at time t>π3B0α would be

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a

32v0B0α+32v0t−π3B0α

b

32v0B0α+v0t−π3B0α

c

32v0B0α+v02t−π3B0α

d

32v0B0α+v0t2

answer is C.

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Detailed Solution

[Step 1]:   r=mV0B0q=V0B0α[Step 2]:   x r=32=sinθ∴θ=600----(1)[Step 3]:   t0A=T6=2πmqB06=π3B0α∵qm=α[Concept]: Therefore x – coordinate of particle at any time  t>π3B0αwill  bestep 4:x'=x0+v0 cosθt−t0A⇒x'=32v0B0α+v0t−π3B0αcos600  ⇒x'=32v0B0α+v02t−π3B0α
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A particle of charge per unit mass α is released from origin with velocity V→=v0i^ in a magnetic field B→=−B0k^  for  x≤32v0B0α  and B→=0  for x>32v0B0α. Then x–coordinate of the particle at time t>π3B0α would be