A beam of positively charged particles is moving with a velocity of...

For the positively charged particles to pass through the region without any deviation, the net force acting on them must be zero. This means the electric force and the magnetic force must be equal and opposite.

The Lorentz force equation is given by:
$\vec{F} = q\vec{E} + q(\vec{v} \times \vec{B}) = 0$

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

Given:
Velocity, $\vec{v} = 4 \times 10^5 \hat{i} \text{ m s}^{-1}$
Electric field, $\vec{E} = 8 \times 10^4 \hat{j} \text{ V m}^{-1}$

Substituting these values into the equation:
$8 \times 10^4 \hat{j} = -(4 \times 10^5 \hat{i} \times \vec{B})$

For the cross product $\hat{i} \times \vec{B}$ to result in a $-\hat{j}$ direction, $\vec{B}$ must be in the $\hat{k}$ direction (since $\hat{i} \times \hat{k} = -\hat{j}$). Let $\vec{B} = B \hat{k}$.

$8 \times 10^4 \hat{j} = -[4 \times 10^5 \hat{i} \times (B \hat{k})]$

$8 \times 10^4 \hat{j} = -[4 \times 10^5 B (-\hat{j})]$

$8 \times 10^4 \hat{j} = 4 \times 10^5 B \hat{j}$

Equating the magnitudes:
$B = \dfrac{8 \times 10^4}{4 \times 10^5} = 0.2 \text{ T}$

Thus, the magnetic field must have a magnitude of $0.2 \text{ T}$ and be directed along the positive z-axis ($\hat{k}$).

Answer: $0.2 \mathrm{~T}$ along the positive z-axis