# How will a positively charged particle move when its velocity is in the direction of the electric field but perpendicular to the magnetic field?

This case is very hard for me to visualise. Like say the particle moves along the +x direction in the direction of the electric field and the magnetic field is along the +y direction then, the forces on it are: due to the magnetic field is upwards initially (+z direction) and along +x due to the electric field and they don't affect each other that's okay i understand that but the next instant where now it has moved in the xz plane the magnetic force will have a component in the -x direction opposing the electric field so now how do we analyse this? Is it possible to do this without any values for the magnitude like the direction the particle will move in now will be determined by what force is greater. am I understanding it wrong?

You can tackle it with Newton's laws along with the work-energy theorem. The force on the particle is $$\vec{F} = m\vec{a} = q(\vec{E} + \vec{v} \times \vec{B)}.$$ If you resolve that into components you get $$\begin{eqnarray} &m\dot{v_x} &= qE + qv_zB \\ &m\dot{v_z} &= qv_xB \end{eqnarray}$$ You can solve the coupled differential equations to obtain $$v_x$$ & $$v_z$$ and integrate them to get the trajectories. I'll leave that to you. Note that there will be no forces in the $$y$$-direction so, by momentum conservation, $$v_y(t) = 0$$.
The exact shape of the trajectory would depend on the values of the parameters $$q,E,B,$$ and $$m$$. But if you notice the differential equation you obtain for $$v_z$$, $$\begin{eqnarray} \ddot{v_z} + \frac{q^2 B^2}{m^2} v_z = \frac{q^2 EB}{m^2}, \end{eqnarray}$$
you'll see that it's a type of oscillatory motion. Also, it takes place in the $$xz-$$plane, since the electrostatic force propels the particles in the $$x-$$direction and the magnetostatic force is always perpendicular to $$y$$.