A particle having a charge of -2 C is moving with a velocity v =...

The magnetic force on a moving charge is given by $\vec{F} = q(\vec{v} \times \vec{B})$.

Substituting the given vectors:

$\vec{v} \times \vec{B} = (v_x \hat{i} + 3 \hat{j}) \times (2 \hat{i} - 4 \hat{j} + B_z \hat{k})$

$\vec{v} \times \vec{B} = 3 B_z \hat{i} - v_x B_z \hat{j} + (-4 v_x - 6) \hat{k}$

Multiplying by the charge $q = -2\text{ C}$:

$\vec{F} = -2 [3 B_z \hat{i} - v_x B_z \hat{j} + (-4 v_x - 6) \hat{k}]$

$\vec{F} = -6 B_z \hat{i} + 2 v_x B_z \hat{j} + (8 v_x + 12) \hat{k}$

Equating this to the given force $\vec{F} = 24 \hat{i} + 16 \hat{j} - 4 \hat{k}$:

For the $\hat{i}$ component:

$-6 B_z = 24 \Rightarrow B_z = -4\text{ T}$

For the $\hat{j}$ component:

$2 v_x B_z = 16$

Substituting $B_z = -4\text{ T}$:

$2 v_x (-4) = 16 \Rightarrow -8 v_x = 16 \Rightarrow v_x = -2\text{ m/s}$

Checking the $\hat{k}$ component:

$8 v_x + 12 = 8(-2) + 12 = -4$, which matches the given force.

Thus, $v_x = -2\text{ m/s}$ and $B_z = -4\text{ T}$.

Answer: $v_x = -2\text{ m/s},\; B_z = -4\text{ T}$