An electron is launched with velocity v→ in a uniform magnetic field B→. The angle θ between V→and B→lies between 0 and π2. Its velocity vector v→ returns to its initial value in a time interval of

depends upon angle between v → and B →

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An electron is launched with velocity $\vec{v}$ in a uniform magnetic field $\vec{B}$ . The angle $θ$ between $\vec{V}$ and $\vec{B}$ lies between 0 and $\frac{π}{2}$ . Its velocity vector $\vec{v}$ returns to its initial value in a time interval of

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$\frac{2 π m}{e B}$

$\frac{2 \times 2 π m}{e B}$

$\frac{π m}{e B}$

depends upon angle between $\vec{v} and \vec{B}$

answer is A.

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Time interval in which $\vec{v}$ returns to its initial value is same as time period of the particle, hence the required time $= \frac{2 π m}{e B}$

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