A particle of mass m and charge q is released from rest at the origin as shown in the figure. The speed of the particle when it has travelled a distance d along the z-axis is given by

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$\sqrt{\frac{2 q E_{0} d}{m}}$

$q \sqrt{\frac{E_{0} d}{B_{0}}}$

$\sqrt{\frac{m}{q E_{0} d}}$

$\sqrt{\frac{2 m}{q E_{0} d}}$

answer is A.

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Detailed Solution

The gain in kinetic energy after travelling through a distance of d along z-axis will be qE₀d, Since work done by Magnetic field will be zero.

$\Rightarrow \frac{1}{2} m v^{2} = q E_{0} d \Rightarrow v = \sqrt{\frac{2 q E_{0} d}{m}}$

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