A particle of specific charge α (charge per unit mass) is released at time t = 0 from origin with an initial velocity of v→=v0i∧ in a uniform magnetic field B→=−B0k∧. Find position of particle at any time t.

θ=wt=2πTt=2π2πmB0qt=B0αt; R=mvBq=V0B0αOP→=Rsin⁡θi^+(R−Rcos⁡θ)j∧=R[sin⁡θi+(1−cos⁡θ)j∧]=V0B0αsin⁡B0αti+1−cos⁡B0αtj∧

A particle of specific charge α (charge per unit mass) is released at time t = 0 from origin with an initial velocity of v→=v0i∧ in a uniform magnetic field B→=−B0k∧. Find position of particle at any time t.

θ=wt=2πTt=2π2πmB0qt=B0αt; R=mvBq=V0B0αOP→=Rsin⁡θi^+(R−Rcos⁡θ)j∧=R[sin⁡θi+(1−cos⁡θ)j∧]=V0B0αsin⁡B0αti+1−cos⁡B0αtj∧

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A particle of specific charge $α$ (charge per unit mass) is released at time t = 0 from origin with an initial velocity of $\vec{v} = v_{0} \overset{\land}{i}$ in a uniform magnetic field $\vec{B} = - B_{0} \overset{\land}{k}$ . Find position of particle at any time t.

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$\frac{v_{0}}{B_{0} α} [Sin α (\hat{i} - {1 + Cos (αt)} \hat{j}]$

$\frac{v_{0}}{B_{0} α} [Sin (B_{0} αt) \hat{i} - [1 + Cos (B_{0} αt)] \hat{j}]$

$\frac{v_{0}}{B_{0} α} [Cos B_{0} αt + \{1 - Sin (B_{0} αt)\} \hat{j}]$

$\frac{v_{0}}{B_{0} α} [Sin (B_{0} αt) \hat{i} + \{1 - Cos (B_{0} αt)\} \hat{j}]$

answer is B.

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Detailed Solution

$θ = wt = \frac{2 π}{T} t = (\frac{2 π}{2 πm} B_{0} q) t$

$= B_{0} αt; R = \frac{mv}{Bq} = \frac{V_{0}}{B_{0} α}$

$\vec{OP} = Rsin θ \hat{i} + (R - Rcos θ) \overset{\land}{j}$

$= R [\sin θi + (1 - \cos θ) \overset{\land}{j}]$

$= \frac{V_{0}}{B_{0} α} [\sin (B_{0} αt) i + (1 - \cos B_{0} αt) \overset{\land}{j}$

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A particle of specific charge α (charge per unit mass) is released at time t = 0 from origin with an initial velocity of v→=v0i∧ in a uniform magnetic field B→=−B0k∧. Find position of particle at any time t.

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A particle of specific charge $α$ (charge per unit mass) is released at time t = 0 from origin with an initial velocity of $\vec{v} = v_{0} \overset{\land}{i}$ in a uniform magnetic field $\vec{B} = - B_{0} \overset{\land}{k}$ . Find position of particle at any time t.