A particle of charged `1 muC` is at rest in a magnetic field `vecB=-2hatk` tesla. Magnetic Lorentz force on the charge particle with respect to an observer moving with velocity `vecv=-6hatims^(-1)` will be

Options

A. Zero
B. `-10^(5)hatjN`
C. `-10^(6)hatJ`
D. `+10^(-5)hatjN`

Answer

To solve the problem, we need to calculate the magnetic Lorentz force acting on a charged particle in a magnetic field, considering the motion of an observer. The Lorentz force is given by the equation:

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

where:
- \( \vec{F} \) is the magnetic Lorentz force,
- \( q \) is the charge of the particle,
- \( \vec{v} \) is the velocity of the particle with respect to the observer,
- \( \vec{B} \) is the magnetic field.

### Step-by-step Solution:

1. **Identify the given values**:
- Charge \( q = 1 \mu C = 1 \times 10^{-6} C \)
- Magnetic field \( \vec{B} = -2 \hat{k} \, \text{T} \)
- Velocity of the observer \( \vec{v}_{\text{observer}} = -6 \hat{i} \, \text{m/s} \)
- Velocity of the charged particle \( \vec{v}_{\text{particle}} = 0 \, \text{m/s} \) (since it is at rest)

2. **Calculate the relative velocity of the charged particle with respect to the observer**:
\[ \vec{v}_{\text{relative}} = \vec{v}_{\text{particle}} - \vec{v}_{\text{observer}} = 0 - (-6 \hat{i}) = 6 \hat{i} \, \text{m/s} \]

3. **Substitute the values into the Lorentz force equation**:
\[ \vec{F} = q (\vec{v}_{\text{relative}} \times \vec{B}) = (1 \times 10^{-6}) (6 \hat{i} \times (-2 \hat{k})) \]

4. **Calculate the cross product \( \hat{i} \times \hat{k} \)**:
- Using the right-hand rule, \( \hat{i} \times \hat{k} = \hat{j} \).
- Therefore, \( 6 \hat{i} \times (-2 \hat{k}) = -12 \hat{j} \).

5. **Substitute back into the force equation**:
\[ \vec{F} = (1 \times 10^{-6}) (-12 \hat{j}) = -12 \times 10^{-6} \hat{j} \, \text{N} \]

6. **Final result**:
\[ \vec{F} = 12 \times 10^{-6} \hat{j} \, \text{N} \]

### Conclusion:
The magnetic Lorentz force on the charged particle with respect to the observer moving with velocity \( \vec{v} = -6 \hat{i} \, \text{m/s} \) is:

\[ \vec{F} = 12 \times 10^{-6} \hat{j} \, \text{N} \]