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Byju's Answer

Standard XII

Physics

Analysis of Force Equation

The particle ...

Question

The particle of charge $q$ and mass $m$ starts moving from origin under the action of an electric field $E = E_{0} i$ and magnetic field $V = V_{0} i$ . Its velocity at $(x, 3, 0)$ is $(4 i + 3^i)$ . The value of $x$ is :

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

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\frac{25 m}{2 q E_{o}}

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\frac{10 m}{q E_{o}}

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\frac{25 E_{0} B_{0}}{m}

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Solution

The correct option is A $\frac{\sqrt{3} m v_{0}}{q E}$
Final velocity of the particle
$\begin{matrix} V = \sqrt{V_{0}^{2} + {(\frac{q E t}{m})}^{2}} = 2 V_{0} V_{0}^{2} + {(\frac{q E t}{m})}^{2} = 4 V_{0}^{2} {(\frac{q E t}{m})}^{2} = 3 V_{0}^{2} t = \frac{\sqrt{3} m v_{0}}{q E} \end{matrix}$

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The particle of charge q and mass m starts moving from origin under the action of an electric field E=E0i and magnetic field V=V0i . Its velocity at (x,3,0) is (4i+3^i) . The value of x is :

The correct option is A √3mv0qEFinal velocity of the particleV=√V20+(qEtm)2=2V0V20+(qEtm)2=4V20(qEtm)2=3V20t=√3mv0qE

The particle of charge $q$ and mass $m$ starts moving from origin under the action of an electric field $E = E_{0} i$ and magnetic field $V = V_{0} i$ . Its velocity at $(x, 3, 0)$ is $(4 i + 3^i)$ . The value of $x$ is :

The correct option is A $\frac{\sqrt{3} m v_{0}}{q E}$
Final velocity of the particle
$\begin{matrix} V = \sqrt{V_{0}^{2} + {(\frac{q E t}{m})}^{2}} = 2 V_{0} V_{0}^{2} + {(\frac{q E t}{m})}^{2} = 4 V_{0}^{2} {(\frac{q E t}{m})}^{2} = 3 V_{0}^{2} t = \frac{\sqrt{3} m v_{0}}{q E} \end{matrix}$