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. `vecE=-u_(0)B_(0)(hati +hatj+7hatk)`
B. `vecE=-u_(0)B_(0)(3hati -3hatj-4hatk)`
C. `vecE=u_(0)B_(0)(14hatj+7hatk)`
D. `vecE=-u_(0)B_(0)(14hatj + 7hatk)`

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 through a magnetic field \( \vec{B} \). The magnetic field and the velocity of the charge are given, and we will use the Lorentz force equation to find the electric field.

### Step-by-step Solution:

1. **Understanding the Lorentz Force Equation**:
The total force \( \vec{F} \) on a charge \( q \) moving with velocity \( \vec{v} \) in electric field \( \vec{E} \) and magnetic field \( \vec{B} \) is given by:
\[ \vec{F} = q(\vec{E} + \vec{v} \times \vec{B}) \]
Since the charge experiences no force, we have:
\[ \vec{E} + \vec{v} \times \vec{B} = 0 \]
Therefore, we can express the electric field as:
\[ \vec{E} = -\vec{v} \times \vec{B} \]

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

3. **Calculating the Cross Product \( \vec{v} \times \vec{B} \)**:
To find \( \vec{v} \times \vec{B} \), we set up the determinant:
\[ \vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3v_0 & -v_0 & 2v_0 \\ B_0 & 2B_0 & -4B_0 \end{vmatrix} \]

Expanding this determinant:
\[ \vec{v} \times \vec{B} = \hat{i} \begin{vmatrix} -v_0 & 2v_0 \\ 2B_0 & -4B_0 \end{vmatrix} - \hat{j} \begin{vmatrix} 3v_0 & 2v_0 \\ B_0 & -4B_0 \end{vmatrix} + \hat{k} \begin{vmatrix} 3v_0 & -v_0 \\ B_0 & 2B_0 \end{vmatrix} \]

Calculating each of these determinants:
- For \( \hat{i} \):
\[ -v_0(-4B_0) - 2v_0(2B_0) = 4v_0B_0 - 4v_0B_0 = 0 \]
- For \( \hat{j} \):
\[ 3v_0(-4B_0) - 2v_0B_0 = -12v_0B_0 - 2v_0B_0 = -14v_0B_0 \]
- For \( \hat{k} \):
\[ 3v_0(2B_0) - (-v_0B_0) = 6v_0B_0 + v_0B_0 = 7v_0B_0 \]

Thus, we have:
\[ \vec{v} \times \vec{B} = 0\hat{i} + 14v_0B_0\hat{j} + 7v_0B_0\hat{k} \]

4. **Finding the Electric Field**:
Now substituting back into the equation for \( \vec{E} \):
\[ \vec{E} = -\vec{v} \times \vec{B} = -\left(0\hat{i} - 14v_0B_0\hat{j} + 7v_0B_0\hat{k}\right) \]
Therefore,
\[ \vec{E} = 0\hat{i} + 14v_0B_0\hat{j} - 7v_0B_0\hat{k} \]

5. **Final Result**:
The electric field in the region is:
\[ \vec{E} = 14v_0B_0\hat{j} - 7v_0B_0\hat{k} \]