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 velocity of the observer. The magnetic Lorentz force is given by the formula:

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

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 1: Identify the given values
- Charge \( Q = 1 \, \mu C = 1 \times 10^{-6} \, C \)
- Magnetic field \( \vec{B} = -2 \hat{k} \, T \)
- Velocity of the observer \( \vec{v}_{observer} = -6 \hat{i} \, m/s \)

### Step 2: Determine the velocity of the charged particle with respect to the observer
Since the charged particle is at rest, its velocity with respect to the observer moving at \( -6 \hat{i} \) will be:

\[ \vec{v}_{particle} = 0 - (-6 \hat{i}) = 6 \hat{i} \, m/s \]

### Step 3: Calculate the cross product \( \vec{v} \times \vec{B} \)
Now we calculate the cross product \( \vec{v}_{particle} \times \vec{B} \):

\[ \vec{v}_{particle} = 6 \hat{i} \]
\[ \vec{B} = -2 \hat{k} \]

Using the right-hand rule for the cross product:

\[ \vec{v}_{particle} \times \vec{B} = (6 \hat{i}) \times (-2 \hat{k}) \]

Using the property of cross products:
\[ \hat{i} \times \hat{k} = \hat{j} \]

So,
\[ \vec{v}_{particle} \times \vec{B} = 6 \times -2 \hat{j} = -12 \hat{j} \]

### Step 4: Calculate the magnetic Lorentz force
Now substituting back into the Lorentz force equation:

\[ \vec{F} = Q (\vec{v} \times \vec{B}) = (1 \times 10^{-6} \, C)(-12 \hat{j}) = -12 \times 10^{-6} \hat{j} \, N \]

### Step 5: Final answer
Thus, the magnetic Lorentz force on the charged particle is:

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

This can also be expressed as:

\[ \vec{F} = -1.2 \times 10^{-5} \hat{j} \, N \]

### Summary of the solution:
The magnetic Lorentz force on the charged particle with respect to an observer moving with a velocity of \( -6 \hat{i} \, m/s \) is \( -1.2 \times 10^{-5} \hat{j} \, N \).

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