The force acting on a charged particle due to electric and magnetic...

Given $\vec{E} = E \hat{i}$ and $\vec{B} = B \hat{j}$.

The net force on the particle is zero, so $\vec{F} = q\vec{E} + q(\vec{v} \times \vec{B}) = 0$.

$\vec{E} = -(\vec{v} \times \vec{B})$

Let $\vec{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k}$.

$\vec{v} \times \vec{B} = (v_x \hat{i} + v_y \hat{j} + v_z \hat{k}) \times (B \hat{j}) = v_x B \hat{k} - v_z B \hat{i}$

Substituting this into the force equation:

$E \hat{i} - v_z B \hat{i} + v_x B \hat{k} = 0$

Comparing the components, we get:

$E - v_z B = 0 \Rightarrow v_z = \dfrac{E}{B}$

$v_x B = 0 \Rightarrow v_x = 0$

The component $v_y$ can be any real number. The speed of the particle is $v = \sqrt{v_x^2 + v_y^2 + v_z^2} = \sqrt{v_y^2 + \left(\dfrac{E}{B}\right)^2}$.

For the minimum speed, $v_y = 0$.

Thus, $\vec{v} = \dfrac{E}{B} \hat{k}$.

The particle must be projected along the positive Z-axis with a minimum speed of $\dfrac{E}{B}$.