Q.

A particle of charge per unit mass α is released from origin with velocity v→=v→0i^ in a magnetic fieldB→=−B0k^  for  x   ≤   32   v0B0αand  B→= 0   for   x  >  32   v0B0αThe x-coordinates of the particle at time t >  π3B0α would be

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a

32  v0B0α  + 32 v0 t−  πB0α

b

32  v0B0α  +  v0 t−  πB0α

c

32  v0B0α  + v02  t−  π3B0α

d

32  v0B0α  +  v0t2

answer is C.

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Detailed Solution

r=mv0B0q  =  v0B0αxr  =  32  =  sin θ   ∴     θ =  600tQA=T6  =    π3B0αTherefore x-coordinate of particle at any time t>  π3B0α  will bex= 32   v0B0α  + v0  t−  π3B0α  cos600=32v0B0α  +  v02  t−π3B0α
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