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To solve this Electrodynamics problem step-by-step:

### Part (a)
We are tasked to determine the DC component (\(v_{\text{DC}}\)) of the velocity \(\mathbf{v}\) of a charged particle in constant electric and magnetic fields (\(\mathbf{E}\) and \(\mathbf{B}\)).

Given:
- Newton's second law is \(\mathbf{F} = m\mathbf{a}\), but for a charged particle, it becomes \(m\mathbf{a} = q\left(\mathbf{E} + \mathbf{v} \times \mathbf{B}\right)\).
- The DC component of the velocity \(\mathbf{v}_{\text{DC}}\) satisfies the time-averaged equation, meaning that \( \mathbf{v}_{\text{AC}} = 0\) over time. 
- Therefore, \( \mathbf{E} + \mathbf{v}_{\text{DC}} \times \mathbf{B} = 0 \).

#### Step 1: Start from the force balance
For the DC component, we know:
\[
\mathbf{E} + \mathbf{v}_{\text{DC}} \times \mathbf{B} = 0
\]
which can be rewritten as:
\[
\mathbf{v}_{\text{DC}} \times \mathbf{B} = -\mathbf{E}
\]
Thus, \(\mathbf{v}_{\text{DC}}\) must be perpendicular to both \(\mathbf{E}\) and \(\mathbf{B}\).

#### Step 2: Use the vector identity
We can solve for \(\mathbf{v}_{\text{DC}}\) by using the vector identity given in the problem. Rearranging and using the formula:
\[
\mathbf{A} \times (\mathbf{B} \times \mathbf{C}) = (\mathbf{A} \cdot \mathbf{C})\mathbf{B} - (\mathbf{A} \cdot \mathbf{B})\mathbf{C}
\]
we get:
\[
\mathbf{v}_{\text{DC}} = \frac{\mathbf{E} \times \mathbf{B}}{B^2}
\]
where \(B^2 = \mathbf{B} \cdot \mathbf{B}\).

### Part (b)
Now, we need to explicitly calculate \(\mathbf{v}_{\text{DC}}\) given the following values:
- \(\mathbf{E} = 5\hat{x} \, \text{mV/m} = 5 \times 10^{-3} \hat{x} \, \text{V/m}\)
- \(\mathbf{B} = 25\hat{y} \, \mu\text{T} = 25 \times 10^{-6} \hat{y} \, \text{T}\)

We use the formula derived above:
\[
\mathbf{v}_{\text{DC}} = \frac{\mathbf{E} \times \mathbf{B}}{B^2}
\]

#### Step 1: Compute the cross product \(\mathbf{E} \times \mathbf{B}\)
\[
\mathbf{E} \times \mathbf{B} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
5 \times 10^{-3} & 0 & 0 \\
0 & 25 \times 10^{-6} & 0
\end{vmatrix} = (0\hat{i} + 0\hat{j} + (5 \times 10^{-3})(25 \times 10^{-6})\hat{k}) = 1.25 \times 10^{-7} \hat{z}
\]

#### Step 2: Compute \(B^2\)
\[
B^2 = (25 \times 10^{-6})^2 = 6.25 \times 10^{-10} \, \text{T}^2
\]

#### Step 3: Calculate \(\mathbf{v}_{\text{DC}}\)
\[
\mathbf{v}_{\text{DC}} = \frac{1.25 \times 10^{-7} \hat{z}}{6.25 \times 10^{-10}} = 200 \, \hat{z} \, \text{m/s}
\]

Thus, the DC component of the velocity is:
\[
\mathbf{v}_{\text{DC}} = 200 \, \hat{z} \, \text{m/s}
\]

### Conclusion:
- The DC velocity component \(\mathbf{v}_{\text{DC}}\) is \(200 \, \hat{z} \, \text{m/s}\), directed along the positive \(z\)-axis.

---

Do you need more details or have any questions? Here are some related questions for further exploration:

1. How does the AC component of velocity affect the overall motion of the charged particle?
2. What happens if the electric or magnetic field is time-varying instead of constant?
3. How does the motion of the particle change if \(\mathbf{E}\) and \(\mathbf{B}\) are not perpendicular?
4. Can we use this same method to solve for \(\mathbf{v}_{\text{DC}}\) in non-uniform fields?
5. What are some real-world applications of this scenario in plasma physics or space science?

**Tip**: When dealing with vector cross products, always ensure the fields are expressed in consistent units and directions before calculating.

Consider Newton’s 2nd Law with F = q(E + v × B) and determine the DC component of the velocity v of the particle with zero component parallel to B. Explicitly solve for v_DC given E = 5x mV/m and B = 25y μT.

This problem involves calculating the DC component of velocity for a charged particle moving through constant electric and magnetic fields using Newton's second law. It includes vector calculus and real-world applications in Earth's ionized atmosphere.

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