A particle having mass $m$ and charge $q$ is released from the origin in a region in which electric field and magnetic fields are given by $\to B = - B_{0}^j$ and $\to E = E_{0}^k$ . Find the speed of the particle as a function of its $z -$ coordinate.

\sqrt{\frac{q E z}{m}}

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\sqrt{\frac{2 (q v B + q E) z}{m}}

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\sqrt{\frac{(- q v B + q E) 2 z}{m}}

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\sqrt{\frac{2 q E z}{m}}

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Solution

The correct option is C $\sqrt{\frac{2 q E z}{m}}$
Given:
$\to B = B_{0}^j \to E = E_{0}^k$
The particle have initial velocity zero, So electric field will be responsible for to move in opposite direction of electric field.
Magnetic force on the particle does not increase the speed of the particle since ${\to F}_{m a g} . \to v = 0$ ( $∠ 90^{\circ}$ )

${\to F}_{e l e c t r i c} = q E^k$
$\Rightarrow \to a = \frac{q E}{m}^k$
Since acceleration is constant applying the one-dimensional kinematics equation:
$v^{2} - u^{2} = 2 a_{z}$
$∴ v^{2} = \frac{2 q E z}{m}$ since initial speed $u = 0$
$\Rightarrow v = \sqrt{\frac{2 q E z}{m}}$

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A particle having mass m and charge q is released from the origin in a region in which electric field and magnetic fields are given by →B=−B0^j and →E=E0^k. Find the speed of the particle as a function of its z−coordinate.

A particle having mass $m$ and charge $q$ is released from the origin in a region in which electric field and magnetic fields are given by $\to B = - B_{0}^j$ and $\to E = E_{0}^k$ . Find the speed of the particle as a function of its $z -$ coordinate.