A particle of charge q and mass m initially at rest- starts moving from the origin under the action of an electric field E-E0- and a magnetic field B- B 0- with velocity v - v 0- The speed of the particle will become 2 v 0 after a timeA- t-2 m v0-a EB- t -2 B - mC- t -3 mv 0- aED- t-3 B q-m v0

The correct option is C t=√(3)mv_0/qEThe E – field accelerates the particle along x – axis with a=qE_0/m⇒ V_x=qE_0t/m Net velocity is, V⃗=qE_0t/mÎ+v_0 Ĵ V^2 =q ...

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Byju's Answer

Standard XII

Physics

Acceleration

A particle of...

Question

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

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

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

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

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

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Solution

The correct option is D $t = \frac{\sqrt{3} m v_{0}}{q E}$
The E – field accelerates the particle along x – axis
$w i t h a = \frac{q E_{0}}{m} \Rightarrow V_{x} = \frac{q E_{0} t}{m}$
$N e t v e l o c i t y i s, \to V = \frac{q E_{0} t}{m}^I + v_{0}^J$
$V^{2} = \frac{q^{2} E_{0}^{2} t^{2}}{m^{2}} \Rightarrow t = \frac{3 \sqrt{3} m V_{0}}{q E_{0}}$

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

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