An electron moving with a velocity `oversetrarrV_(1) = 2hatim//s` at a point in a magnetic field experiences a force `oversetrarrF_(1) = 2hatjN` if the elctron is moving with a velocity `oversetrarrV_(2) = 2hatj m//s` at the same point, it experiences a force `oversetrarrF_(2) =+2hatiN` The force the electron wolud experience if it were moving with a velocity `oversetrarrV_(3)=2hatkm//s` at the same point is .

Options

A. (a)zero
B. (b)`2hatkN`
C. (c)`-2hatkN`
D. (d)Information is insufficient

Explanation

To solve the problem, we will use the relationship between the magnetic force experienced by a charged particle moving in a magnetic field, which is given by the equation:

\[ \vec{F} = q (\vec{V} \times \vec{B}) \]

where:
- \(\vec{F}\) is the magnetic force,
- \(q\) is the charge of the particle (for an electron, \(q = -e\)),
- \(\vec{V}\) is the velocity of the particle,
- \(\vec{B}\) is the magnetic field.

### Step 1: Analyze the given information

1. **First case**:
- Velocity: \(\vec{V_1} = -2 \hat{i} \, \text{m/s}\)
- Force: \(\vec{F_1} = 2 \hat{j} \, \text{N}\)

2. **Second case**:
- Velocity: \(\vec{V_2} = 2 \hat{j} \, \text{m/s}\)
- Force: \(\vec{F_2} = 2 \hat{i} \, \text{N}\)

### Step 2: Set up the equations for the magnetic field

Using the force equation for both cases, we can express the magnetic field \(\vec{B}\) in terms of the known velocities and forces.

For the first case:
\[ \vec{F_1} = q (\vec{V_1} \times \vec{B}) \]
Substituting the values:
\[ 2 \hat{j} = -e (-2 \hat{i} \times \vec{B}) \]
This simplifies to:
\[ 2 \hat{j} = 2e (\hat{i} \times \vec{B}) \]
Thus:
\[ \hat{j} = e (\hat{i} \times \vec{B}) \quad \text{(1)} \]

For the second case:
\[ \vec{F_2} = q (\vec{V_2} \times \vec{B}) \]
Substituting the values:
\[ 2 \hat{i} = -e (2 \hat{j} \times \vec{B}) \]
This simplifies to:
\[ 2 \hat{i} = -2e (\hat{j} \times \vec{B}) \]
Thus:
\[ -\hat{i} = e (\hat{j} \times \vec{B}) \quad \text{(2)} \]

### Step 3: Solve for the magnetic field \(\vec{B}\)

From equations (1) and (2), we can find the components of \(\vec{B}\).

1. From equation (1):
- If \(\hat{j} = e (\hat{i} \times \vec{B})\), we can deduce that \(\vec{B}\) has components in the \(k\) direction.

2. From equation (2):
- If \(-\hat{i} = e (\hat{j} \times \vec{B})\), we can also deduce that \(\vec{B}\) has components in the \(k\) direction.

Thus, we can conclude that:
\[ \vec{B} = B_k \hat{k} \]

### Step 4: Calculate the force for the third velocity

Now, we need to find the force when the electron is moving with:
\[ \vec{V_3} = 2 \hat{k} \, \text{m/s} \]

Using the force equation:
\[ \vec{F_3} = q (\vec{V_3} \times \vec{B}) \]
Substituting the values:
\[ \vec{F_3} = -e (2 \hat{k} \times B_k \hat{k}) \]
Since \(\hat{k} \times \hat{k} = 0\), we have:
\[ \vec{F_3} = 0 \]

### Final Answer
The force the electron would experience if it were moving with a velocity \(\vec{V_3} = 2 \hat{k} \, \text{m/s}\) at the same point is:
\[ \vec{F_3} = 0 \, \text{N} \]