The magnetic force is given by:
\[ \vec{F} = q (\vec{v} \times \vec{B}) \]
Substituting the values:
\[ 5e \hat{k} = e \left( 3\hat{i} + 5\hat{j} \right) \times \left( B_0 \hat{i} + 2B_0 \hat{j} \right) \]
Calculating the cross product:
\[ 5e \hat{k} = e \left( 6B_0 \hat{k} - 5B_0 \hat{k} \right) \]
Simplifying:
\[ 5e \hat{k} = e B_0 \hat{k} \]
Therefore:
\[ B_0 = 5T \]
Let $ f: \mathbb{R} \to \mathbb{R} $ be a twice differentiable function such that $$ f''(x)\sin\left(\frac{x}{2}\right) + f'(2x - 2y) = (\cos x)\sin(y + 2x) + f(2x - 2y) $$ for all $ x, y \in \mathbb{R} $. If $ f(0) = 1 $, then the value of $ 24f^{(4)}\left(\frac{5\pi}{3}\right) $ is: