Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

The magnetic force acting on a charged particle of charge $-2 \mu C$ in a magnetic field of $2 T$ acting in y-direction, when the particle velocity is $(2\hat{i} + 3\hat{j}) \times 10^6 \text{ ms}^{-1}$ is:

8 N in -z-direction

4 N in z-direction

8 N in y-direction

4 N in y-direction

Step-by-Step Solution

Formula: The magnetic force $\vec{F}$ exerted on a charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ is given by the Lorentz force equation: $\vec{F} = q(\vec{v} \times \vec{B})$ .
Given Values:

Charge $q = -2 \mu\text{C} = -2 \times 10^{-6} \text{ C}$
Velocity $\vec{v} = (2\hat{i} + 3\hat{j}) \times 10^6 \text{ ms}^{-1}$
Magnetic Field $\vec{B} = 2\hat{j} \text{ T}$

Calculation:

First, calculate the cross product $\vec{v} \times \vec{B}$ : $\vec{v} \times \vec{B} = [(2\hat{i} + 3\hat{j}) \times 10^6] \times (2\hat{j})$ $= 10^6 [ (2\hat{i} \times 2\hat{j}) + (3\hat{j} \times 2\hat{j}) ]$
Using unit vector properties ( $\hat{i} \times \hat{j} = \hat{k}$ and $\hat{j} \times \hat{j} = 0$ ): $\vec{v} \times \vec{B} = 10^6 (4\hat{k} + 0) = 4 \times 10^6 \hat{k}$
Now, multiply by the charge $q$ : $\vec{F} = (-2 \times 10^{-6}) \times (4 \times 10^6 \hat{k})$ $\vec{F} = -8 \hat{k} \text{ N}$

Conclusion: The magnitude of the force is $8 \text{ N}$ , and the negative sign with $\hat{k}$ indicates it acts in the negative z-direction .

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started