A particle of charge q and mass m starts moving from the origin under the action of an electric field E- - E0- and B-B0- with a velocity v- - v0 - The speed of the particle will become 2v0 after a time-

Final velocity of the particle, v=√(v_0^2+(qEtm)^2)=2v_0 v_0^2+(qEtm)^2=4v_0^2 (qEtm)^2=3v_0^2 t=√(3) mv_0qE ...

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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 a velocity $\to v = v_{0}^i$ . The speed of the particle will become $2 v_{0}$ after a time.

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Solution

Final velocity of the particle, $v = \sqrt{v_{0}^{2} + {(\frac{q E t}{m})}^{2}} = 2 v_{0}$
$v_{0}^{2} + {(\frac{q E t}{m})}^{2} = 4 v_{0}^{2}$
${(\frac{q E t}{m})}^{2} = 3 v_{0}^{2}$
$t = \frac{\sqrt{3} m v_{0}}{q E}$

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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 a velocity →v=v0^i. The speed of the particle will become 2v0 after a time.

Final velocity of the particle, v=√v20+(qEtm)2=2v0v20+(qEtm)2=4v20(qEtm)2=3v20t=√3mv0qE

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 a velocity $\to v = v_{0}^i$ . The speed of the particle will become $2 v_{0}$ after a time.

Final velocity of the particle, $v = \sqrt{v_{0}^{2} + {(\frac{q E t}{m})}^{2}} = 2 v_{0}$
$v_{0}^{2} + {(\frac{q E t}{m})}^{2} = 4 v_{0}^{2}$
${(\frac{q E t}{m})}^{2} = 3 v_{0}^{2}$
$t = \frac{\sqrt{3} m v_{0}}{q E}$