A particle of mass m charge q is projected with velocity $v_{0}$ is viscous medium where uniform and constant magnetic field B exists everywhere is the direction perpendicular to the direction of projection of the particle. A viscous force on the particle is given by $\vec{f} = - b \vec{v}$ , where b is constant and $\vec{v}$ is velocity of particle. Choose correct options:

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$\frac{d θ}{d t}$ is given by $\frac{q B}{m}$

Distance travelled by the particle during a time interval form the instant of projection until velocity vector turns by $2 π$ radian is $\frac{m v_{0}}{b} [1 - e^{- \frac{2 π b}{q B}}]$

$\frac{d θ}{d t}$ is time dependent

$\vec{V}$ at any time is $\vec{V} = V_{0} e^{\frac{- b t}{m}} [\cos (\frac{q B}{m} t) \hat{i} + \sin (\frac{q B}{m} t) \hat{j}]$

answer is A, C, D.

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Detailed Solution

$ω = \frac{q B}{m} \frac{d V}{d t} = - \frac{b V}{m} θ = \frac{q B t}{m} V = V_{0} e \frac{- b t}{m} V = V_{0} e^{\frac{- b t}{m}} (\cos w t \hat{i} + \sin w t \hat{j})$

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