A charged particle moving along positive x direction with a velocity V enters a region where there is a uniform magnetic field B→=−B0k^, from x=0 to x=d. If the particle gets deflected. at an angle θ from its initial path, the specific charge (Qm) of the particle is

ω=QBm=αBR=Vω=VαBoNow d=Rsin⁡θ⇒R=dsin⁡θ∴ dsin⁡θ=VαB0⇒α=Vsin⁡θdB0

ω=QBm=αBR=Vω=VαBoNow d=Rsin⁡θ⇒R=dsin⁡θ∴ dsin⁡θ=VαB0⇒α=Vsin⁡θdB0

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A charged particle moving along positive x direction with a velocity V enters a region where there is a uniform magnetic field $\vec{B} = - B_{0} \hat{k}$ , from $x = 0 to x = d$ . If the particle gets deflected. at an angle $θ$ from its initial path, the specific charge $(\frac{Q}{m})$ of the particle is

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$\frac{V \cos θ}{B_{0} d}$

$\frac{V \tan θ}{B_{0} d}$

$\frac{V}{B_{0} d}$

$\frac{V \sin θ}{B_{0} d}$

answer is D.

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$\begin{array}{l} ω = \frac{Q B}{m} = α B \\ R = \frac{V}{ω} = \frac{V}{α B_{o}} \end{array}$

Now $d = R \sin θ \Rightarrow R = \frac{d}{\sin θ}$

$∴ \frac{d}{\sin θ} = \frac{V}{α B_{0}} \Rightarrow α = \frac{V \sin θ}{d B_{0}}$

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