Question

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

• A. \( 50 \hat{i} + 50 \hat{j} \)
• B. \( 100 \hat{i} + 50 \hat{j} \)
• C. \( -50 \hat{i} + 100 \hat{j} \)
• D. \( 50 \hat{i} + 100 \hat{j} \)

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A charged particle moving in both an electric and magnetic field experiences the Lorentz force.
We can express the unknown velocity as a vector in the \(x\)-\(y\) plane (\(\vec{v} = v_x\hat{i} + v_y\hat{j}\)) and substitute it into the Lorentz force equation to solve for its components.
Step 2: Key Formula or Approach:
The Lorentz force equation is \(\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})\).
Given values:
\(q = 10^{-9} \text{ C}\)
\(\vec{E} = 0.4 \hat{j} \text{ N/C}\)
\(\vec{B} = 4 \times 10^{-3} \hat{k} \text{ T}\)
\(\vec{F} = (4\hat{i} + 2\hat{j}) \times 10^{-10} \text{ N}\)
Step 3: Detailed Explanation:
Let the velocity be \(\vec{v} = v_x \hat{i} + v_y \hat{j}\) since the particle is moving in the \(x\)-\(y\) plane.
First, compute the cross product \(\vec{v} \times \vec{B}\):
\[ \vec{v} \times \vec{B} = (v_x \hat{i} + v_y \hat{j}) \times (4 \times 10^{-3} \hat{k}) \] Using the cross product rules \(\hat{i} \times \hat{k} = -\hat{j}\) and \(\hat{j} \times \hat{k} = \hat{i}\):
\[ \vec{v} \times \vec{B} = -v_x(4 \times 10^{-3})\hat{j} + v_y(4 \times 10^{-3})\hat{i} = (4 \times 10^{-3} v_y)\hat{i} - (4 \times 10^{-3} v_x)\hat{j} \] Now substitute everything into the Lorentz force equation divided by \(q\):
\[ \frac{\vec{F}}{q} = \vec{E} + \vec{v} \times \vec{B} \] \[ \frac{10^{-10} (4\hat{i} + 2\hat{j})}{10^{-9}} = 0.4 \hat{j} + (4 \times 10^{-3} v_y)\hat{i} - (4 \times 10^{-3} v_x)\hat{j} \] \[ 0.4 \hat{i} + 0.2 \hat{j} = (4 \times 10^{-3} v_y)\hat{i} + (0.4 - 4 \times 10^{-3} v_x)\hat{j} \] Equate the \(\hat{i}\) components:
\[ 0.4 = 4 \times 10^{-3} v_y \implies v_y = \frac{0.4}{4 \times 10^{-3}} = \frac{400}{4} = 100 \text{ m/s} \] Equate the \(\hat{j}\) components:
\[ 0.2 = 0.4 - 4 \times 10^{-3} v_x \] \[ 4 \times 10^{-3} v_x = 0.4 - 0.2 = 0.2 \] \[ v_x = \frac{0.2}{4 \times 10^{-3}} = \frac{200}{4} = 50 \text{ m/s} \] Thus, the velocity vector is \(\vec{v} = 50\hat{i} + 100\hat{j}\).
Step 4: Final Answer:
The velocity of the particle is \(50\hat{i} + 100\hat{j}\).