In a certain region static electric and magentic fields exist. The magnetic field is given by `vec(B) = B_(0) (hat(i) + 2hat(j)-4hat(k))`. If a test charge moving with a velocity, `vec(upsilon) = upsilon_(0)(3hat(i)-hat(j) +2hat(k))` experience no force in that region, then the electric field in the region, in SI units, is-

Options

A. `vec(E) = -upsilon_(0) B_(0)(hat(i) + hat(j) + 7hat(k))`
B. `vec(E) = -upsilon_(0)B_(0)(3hat(i) - 2hat(j) -4hat(k))`
C. `vec(E) = upsilon_(0)B_(0)(14hat(j) + 7 hat(k))`
D. `vec(E) = -upsilon_(0) B_(0) (14hat(j) + 7hat(k))`

Answer

To solve the problem, we need to find the electric field \(\vec{E}\) in a region where a test charge experiences no force while moving in the presence of static electric and magnetic fields. The magnetic field \(\vec{B}\) and the velocity \(\vec{v}\) of the charge are given.

### Step-by-Step Solution:

1. **Identify the Given Vectors**:
- The magnetic field is given as:
\[ \vec{B} = B_0 (\hat{i} + 2\hat{j} - 4\hat{k}) \]
- The velocity of the test charge is given as:
\[ \vec{v} = v_0 (3\hat{i} - \hat{j} + 2\hat{k}) \]

2. **Use the Lorentz Force Equation**:
The force \(\vec{F}\) on a charge \(q\) moving in electric and magnetic fields is given by:
\[ \vec{F} = q(\vec{E} + \vec{v} \times \vec{B}) \]
Since the test charge experiences no force, we have:
\[ \vec{E} + \vec{v} \times \vec{B} = 0 \]
Therefore:
\[ \vec{E} = -\vec{v} \times \vec{B} \]

3. **Calculate the Cross Product \(\vec{v} \times \vec{B}\)**:
We need to compute the cross product:
\[ \vec{v} \times \vec{B} = v_0 (3\hat{i} - \hat{j} + 2\hat{k}) \times B_0 (\hat{i} + 2\hat{j} - 4\hat{k}) \]
Using the determinant form for the cross product:
\[ \vec{v} \times \vec{B} = v_0 B_0 \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 2 \\ 1 & 2 & -4 \end{vmatrix} \]

4. **Calculate the Determinant**:
Expanding the determinant:
\[ \vec{v} \times \vec{B} = v_0 B_0 \left( \hat{i} \begin{vmatrix} -1 & 2 \\ 2 & -4 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 2 \\ 1 & -4 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & -1 \\ 1 & 2 \end{vmatrix} \right) \]
Calculating each of the 2x2 determinants:
- For \(\hat{i}\):
\[ \begin{vmatrix} -1 & 2 \\ 2 & -4 \end{vmatrix} = (-1)(-4) - (2)(2) = 4 - 4 = 0 \]
- For \(\hat{j}\):
\[ \begin{vmatrix} 3 & 2 \\ 1 & -4 \end{vmatrix} = (3)(-4) - (2)(1) = -12 - 2 = -14 \]
- For \(\hat{k}\):
\[ \begin{vmatrix} 3 & -1 \\ 1 & 2 \end{vmatrix} = (3)(2) - (-1)(1) = 6 + 1 = 7 \]

5. **Combine the Results**:
Thus, we have:
\[ \vec{v} \times \vec{B} = v_0 B_0 \left( 0 \hat{i} + 14 \hat{j} + 7 \hat{k} \right) = v_0 B_0 (0 \hat{i} - 14 \hat{j} + 7 \hat{k}) \]

6. **Find the Electric Field**:
Now substituting back into the equation for \(\vec{E}\):
\[ \vec{E} = -\vec{v} \times \vec{B} = -v_0 B_0 (0 \hat{i} - 14 \hat{j} + 7 \hat{k}) = v_0 B_0 (0 \hat{i} + 14 \hat{j} - 7 \hat{k}) \]

### Final Answer:
The electric field in the region is:
\[ \vec{E} = v_0 B_0 (0 \hat{i} + 14 \hat{j} - 7 \hat{k}) \]