An alpha particle is moving in a magnetic field, $(3\hat i+2\hat j) T$ with the velocity $(7 × 10^5\hat i) m/s$. The magnetic force acting on the particle is

$4.48 × 10^{-13}\hat kN$

The correct answer is Option (2) → $4.48 × 10^{-13}\hat kN$

Magnetic force on a particle: $\vec{F} = q \vec{v} \times \vec{B}$

Given: $\vec{v} = 7 \times 10^5 \hat{i} \, \text{m/s}$, $\vec{B} = 3 \hat{i} + 2 \hat{j} \, \text{T}$, $q = 2e = 2 \times 1.6 \times 10^{-19} \, C$

Cross product: $\vec{v} \times \vec{B} = (7 \times 10^5 \hat{i}) \times (3 \hat{i} + 2 \hat{j}) = (7 \times 10^5 \hat{i}) \times 3 \hat{i} + (7 \times 10^5 \hat{i}) \times 2 \hat{j}$

Note: $\hat{i} \times \hat{i} = 0$, $\hat{i} \times \hat{j} = \hat{k}$

So, $\vec{v} \times \vec{B} = 0 + (7 \times 10^5 \cdot 2) \hat{k} = 1.4 \times 10^6 \hat{k}$

Force: $\vec{F} = q \vec{v} \times \vec{B} = 3.2 \times 10^{-19} \cdot 1.4 \times 10^6 \hat{k}$

$\vec{F} = 4.48 \times 10^{-13} \hat{k} \, \text{N}$

Magnetic force: $\vec{F} = 4.48 \times 10^{-13} \hat{k}$ N