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