A particle of charge $$q$$ and mass $$m$$ is projected  from origin with an initial velocity has $$\vec{v} = \frac{v_0}{\sqrt{2}}\hat{x} + \frac{v_0}{\sqrt{2}}\hat{y}$$. There exists a uniform magnetic field $$\vec{B} = B_0\hat{z}$$ and a space varying electric field $$\vec{E} = E_0 e^{-\lambda x}\hat{x}$$ within the region $$0 \leq x \leq L$$. After travelling a distance such that x-coordinate has changed from $$x=0$$ to $$x=L$$, the change in the kinetic energy  is_________.

$$\Delta K = W_{\text{total}} = W_{\text{electric}} + W_{\text{magnetic}}$$

The magnetic force acting on a moving charge is given by the Lorentz force expression $$\vec{F}_B = q(\vec{v} \times \vec{B})$$. Since $$\vec{F}_B$$ is always perpendicular to the velocity vector $$\vec{v}$$ (and consequently perpendicular to the instantaneous displacement vector $$d\vec{r}$$), it does zero work on the particle:

$$W_{\text{magnetic}} = \int \vec{F}_B \cdot d\vec{r} = 0$$

$$W_{\text{electric}} = \int \vec{F}_E \cdot d\vec{r}$$

$$\Delta K = W_{\text{electric}} = \int_{x=0}^{x=L} (q\vec{E}) \cdot d\vec{r}$$

$$\Delta K = \int_{0}^{L} \left( q E_0 e^{-\lambda x} \hat{i} \right) \cdot \left( dx\hat{i} + dy\hat{j} + dz\hat{k} \right)$$

$$\Delta K = \int_{0}^{L} q E_0 e^{-\lambda x} \, dx$$

$$\Delta K = q E_0 \left[ \frac{e^{-\lambda x}}{-\lambda} \right]_{0}^{L}$$

$$\Delta K = -\frac{q E_0}{\lambda} \left( e^{-\lambda L} - 1 \right)$$

$$\Delta K = \frac{q E_0}{\lambda} \left( 1 - e^{-\lambda L} \right)$$