$$
   \mathbf{B} = \mathbf{T} \times \mathbf{N} = \begin{vmatrix}
   \mathbf{i} & \mathbf{j} & \mathbf{k} \\
   -4 & 0 & 16 \\
   0 & -16 & 0
   \end{vmatrix} = (0 \cdot 0 - 16 \cdot (-16)) \mathbf{i} - (-4 \cdot 0 - 16 \cdot 0) \mathbf{j} + (-4 \cdot (-16) - 0 \cdot 0) \mathbf{k}
$$

Let's compute the cross product step by step:

The cross product $\mathbf{B} = \mathbf{T} \times \mathbf{N}$ is given by the determinant:

\[
\mathbf{B} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-4 & 0 & 16 \\
0 & -16 & 0
\end{vmatrix}
\]

Expanding this determinant:

\[
\mathbf{B} = \mathbf{i} \left( 0 \cdot 0 - 16 \cdot (-16) \right) - \mathbf{j} \left( -4 \cdot 0 - 16 \cdot 0 \right) + \mathbf{k} \left( -4 \cdot (-16) - 0 \cdot 0 \right)
\]

Simplifying each component:

For the $\mathbf{i}$ component:
\[
\mathbf{i} (0 - (-256)) = \mathbf{i} \cdot 256
\]

For the $\mathbf{j}$ component:
\[
- \mathbf{j} (0 - 0) = 0
\]

For the $\mathbf{k}$ component:
\[
\mathbf{k} (64 - 0) = \mathbf{k} \cdot 64
\]

Therefore, the cross product is:

\[
\mathbf{B} = 256\mathbf{i} + 0\mathbf{j} + 64\mathbf{k}
\]

So,

\[
\mathbf{B} = 256\mathbf{i} + 64\mathbf{k}
\]

Would you like any further details or have any questions about this computation?

Here are 5 questions you might consider next:
1. How do you find the magnitude of a vector?
2. What are some applications of the cross product in physics?
3. How do you determine if two vectors are perpendicular?
4. How can you express a vector in polar coordinates?
5. What is the dot product and how is it different from the cross product?

**Tip:** When computing determinants, always remember to carefully handle signs and zeros to avoid common errors.

Learn how to compute the cross product of vectors using determinants step-by-step. This problem involves vector algebra, specifically the cross product calculation using determinant formulas. Suitable for advanced high school students.

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