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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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2πmeB

2×2πmeB

πmeB

depends upon angle between v→ and B→

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detailed solution

Correct option is A

Time interval in which v→ returns to its initial value is same as time period of the particle, hence the required time =2πmeB

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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

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Ready to Test Your Skills?

free study material

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