A particle having charge \(10^{-9}\) C moving in \(x\)-\(y\) plane in fields of \(0.4\hat{j}\) N/C and \(4 \times 10^{-3} \hat{k}\) T experiences a force of \((4\hat{i} + 2\hat{j}) \times 10^{-10}\) N. The velocity of the particle at that instant is _______ m/s.

The Lorentz force experienced by a charged particle is given by \(\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})\).

Given that the particle moves in the \(x\)-\(y\) plane, its velocity can be written as \(\vec{v} = v_x \hat{i} + v_y \hat{j}\).

Substituting the given values into the force equation:

\(\vec{F} = 10^{-9} [0.4\hat{j} + (v_x \hat{i} + v_y \hat{j}) \times (4 \times 10^{-3} \hat{k})]\)

Using the cross product rules \(\hat{i} \times \hat{k} = -\hat{j}\) and \(\hat{j} \times \hat{k} = \hat{i}\):

\(\vec{F} = 10^{-9} [0.4\hat{j} - 4 \times 10^{-3} v_x \hat{j} + 4 \times 10^{-3} v_y \hat{i}]\)

\(\vec{F} = 10^{-9} [4 \times 10^{-3} v_y \hat{i} + (0.4 - 4 \times 10^{-3} v_x) \hat{j}]\)

The given force is \(\vec{F} = (4\hat{i} + 2\hat{j}) \times 10^{-10}\) N, which can be rewritten as:

\(\vec{F} = 10^{-9} (0.4\hat{i} + 0.2\hat{j})\) N

Comparing the coefficients of \(\hat{i}\) and \(\hat{j}\) from both expressions:

For the \(\hat{i}\) component:
\(4 \times 10^{-3} v_y = 0.4\)
\(v_y = \dfrac{0.4}{4 \times 10^{-3}} = 100\) m/s

For the \(\hat{j}\) component:
\(0.4 - 4 \times 10^{-3} v_x = 0.2\)
\(4 \times 10^{-3} v_x = 0.2\)
\(v_x = \dfrac{0.2}{4 \times 10^{-3}} = 50\) m/s

Therefore, the velocity of the particle is \(\vec{v} = 50\hat{i} + 100\hat{j}\) m/s.

Answer: \(50\hat{i} + 100\hat{j}\)