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(C) The work done by the electric field is equal to the change in kinetic energy of the particle.Since the magnetic field does no work on a moving cha

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$\frac{25}{2 \alpha E_0}$

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(C) The work done by the electric field is equal to the change in kinetic energy of the particle.Since the magnetic field does no work on a moving charge,only the electric field contributes to the change in kinetic energy.Work done $W = q E_0 x_0 = \Delta K = \frac{1}{2} m v^2$.Given the specific charge $\alpha = \frac{q}{m}$,we can write $q = m \alpha$.Substituting this into the work-energy equation: $m \alpha E_0 x_0 = \frac{1}{2} m v^2$.Simplifying for $x_0$: $x_0 = \frac{v^2}{2 \alpha E_0}$.The velocity vector is $\vec{v} = 4 \hat{i} + 3 \hat{j}$,so the speed $v = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5$.Substituting $v = 5$ into the expression for $x_0$: $x_0 = \frac{5^2}{2 \alpha E_0} = \frac{25}{2 \alpha E_0}$.

$A$ particle of specific charge (charge/mass) $\alpha$ starts moving from the origin under the action of an electric field $\vec{E} = E_0 \hat{i}$ and a magnetic field $\vec{B} = B_0 \hat{k}$. Its velocity at $(x_0, y_0, 0)$ is $(4 \hat{i} + 3 \hat{j})$. The value of $x_0$ is:

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