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Step-by-Step Solution
Step 1: Identify the Given Information
• Charge of the particle, $q = -2 \times 10^{-6}\,\text{C}$
• Magnetic field, $\vec{B} = 2\,\hat{j}\,\text{T}$
• Velocity of the particle, $\vec{v} = (2\hat{i} + 3\hat{j}) \times 10^6\,\text{m/s}$
Step 2: Write the Formula for Magnetic Force
The magnetic force on a charged particle is given by:
$$
\vec{F} = q \left( \vec{v} \times \vec{B} \right)
$$
Step 3: Calculate the Cross Product $\vec{v} \times \vec{B}$
Given
$$
\vec{v} = (2 \times 10^6)\hat{i} + (3 \times 10^6)\hat{j}, \quad
\vec{B} = 2\hat{j}.
$$
The cross product $\vec{v} \times \vec{B}$ is:
$$
\vec{v} \times \vec{B}
= (2 \times 10^6 \hat{i} + 3 \times 10^6 \hat{j}) \times (2 \hat{j}).
$$
We break this into two parts:
1. $(2 \times 10^6 \hat{i}) \times (2 \hat{j}) = 4 \times 10^6 (\hat{i} \times \hat{j})$.
Since $\hat{i} \times \hat{j} = \hat{k}$,
$$
\implies 4 \times 10^6 \hat{k}.
$$
2. $(3 \times 10^6 \hat{j}) \times (2 \hat{j}) = 0,$
because $\hat{j} \times \hat{j} = \vec{0}$.
Hence,
$$
\vec{v} \times \vec{B} = 4 \times 10^6 \hat{k}.
$$
Step 4: Multiply by the Charge $q$
Now, we multiply this cross product by the charge $q = -2 \times 10^{-6}\,\text{C}$:
$$
\vec{F} = \left( -2 \times 10^{-6} \right) \times \left( 4 \times 10^6 \hat{k} \right).
$$
Simplifying:
$$
\vec{F} = -8 \hat{k}\,\text{N}.
$$
This indicates a force of magnitude $8\,\text{N}$ in the negative $z$-direction.
Step 5: State the Final Answer
The magnitude of the magnetic force acting on the charged particle is $8\,\text{N}$ and it points along the negative $z$-axis. Therefore, the correct answer is:
$$
\text{8 N in } -z \text{ direction}.
$$
