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In a certain region of space, there exists a uniform and constant electric field of strength E along x-axis and uniform constant magnetic field of induction B along z-axis. A charged particle having charge q and mass m is projected with speed v parallel to x-axis from a point (a, b, 0). When the particle reaches a point $(2 a, \frac{b}{2}, 0)$ its speed becomes 2v. Find the value of electric field strength.

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$\frac{3}{2} \frac{{mv}^{2}}{qa}$

$\frac{{mv}^{2}}{q B}$

$\frac{2 {mv}^{2}}{q B a}$

$\frac{3}{2} vB$

answer is A.

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Detailed Solution

Change in kinetic energy is only due to work done by electric field. Hence, work done by electric field is change in KE.

$qEa = \frac{1}{2} m [{(2 v)}^{2} - v^{2}] \Rightarrow E = \frac{3 {mv}^{2}}{2 qa}$

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