Question

The magnetic force acting on a charged particle carrying a charge of \( 3 \, \mu C \) in a magnetic field of 5 T acting in the \( y \)-direction, when the particle velocity is \( ( \hat{i} + \hat{j} ) \times 10^5 \, \text{ms}^{-1} \), is:

• A. 0.5 N in \( +x \) direction
• B. 0.2 N in \( +y \) direction
• C. 2 N in \( -x \) direction
• D. 1.5 N in \( +z \) direction
• E. 1 N in \( +z \) direction

Hint:

When calculating the magnetic force, remember that the force is always perpendicular to both the velocity of the particle and the direction of the magnetic field. Use the right-hand rule for cross products.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The magnetic force ($\vec{F}$) on a moving charge is given by the Lorentz force equation. The direction of the force is always perpendicular to both the velocity and the magnetic field vectors.
Step 2: Key Formula or Approach
The vector form of magnetic force is: \[ \vec{F} = q(\vec{v} \times \vec{B}) \]
Step 3: Detailed Calculation
1. Identify the vectors: - $q = 3 \times 10^{-6}$ C - $\vec{v} = (10^5 \hat{i} + 10^5 \hat{j})$ m/s - $\vec{B} = 5 \hat{j}$ T 2. Calculate the cross product \( \vec{v} \times \vec{B} \): - $(10^5 \hat{i} + 10^5 \hat{j}) \times (5 \hat{j})$ - $= (10^5 \times 5)(\hat{i} \times \hat{j}) + (10^5 \times 5)(\hat{j} \times \hat{j})$ - Since $\hat{i} \times \hat{j} = \hat{k}$ and $\hat{j} \times \hat{j} = 0$: - $= 5 \times 10^5 \hat{k}$ 3. Calculate the force \( \vec{F} \): - $\vec{F} = (3 \times 10^{-6}) \times (5 \times 10^5 \hat{k})$ - $\vec{F} = 15 \times 10^{-1} \hat{k} = 1.5 \hat{k}$ N
Step 4: Final Answer
The force is 1.5 N in the +z direction.