A particle of charge q and mass m starts moving from the origin under the action of an electric field E- - E 0 - and B- - B 0 - with velocity v- - v 0 - The speed of the particle becomes 2v0 after a time of t-

The correct option is D √( 3 ) m v _ 0 qE undefined ...

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Byju's Answer

Standard XII

Physics

Cyclotron

A particle of...

Question

A particle of charge $q$ and mass $m$ starts moving from the origin under the action of an electric field $\to E = E_{0}^i$ and $\to B = B_{0}^i$ with velocity $\to v = v_{0}^j$ . The speed of the particle becomes $2 v_{0}$ after a time of t=,

\frac{2 m v_{0}}{q E}

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\frac{2 B q}{m v_{0}}

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\frac{\sqrt{3} B q}{m v_{0}}

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\frac{\sqrt{3} m v_{0}}{q E}

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Solution

The correct option is D $\frac{\sqrt{3} m v_{0}}{q E}$
undefined

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A particle of charge q and mass m starts moving from the origin under the action of an electric field →E=E0^i and →B=B0^i with velocity →v=v0^j. The speed of the particle becomes 2v0 after a time of t=,

The correct option is D √3mv0qEundefined

A particle of charge $q$ and mass $m$ starts moving from the origin under the action of an electric field $\to E = E_{0}^i$ and $\to B = B_{0}^i$ with velocity $\to v = v_{0}^j$ . The speed of the particle becomes $2 v_{0}$ after a time of t=,

The correct option is D $\frac{\sqrt{3} m v_{0}}{q E}$
undefined