Question

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 is $ (2 \hat{i} + 3\hat{j}) \times 10^6 \, ms^{-1}$

• A. 4 N in z direction
• B. 8 N in y direction
• C. 8 N in z direction
• D. 8 N in - z direction

Answer 1

The Correct Option is D

 To solve this problem, we need to determine the magnetic force acting on a charged particle. The magnetic force experienced by a charged particle in a magnetic field is given by the Lorentz force formula:

\(F = q (\vec{v} \times \vec{B})\)

Where:

• \(F\) is the magnetic force
• \(q\) is the charge of the particle
• \(\vec{v}\) is the velocity of the particle
• \(\vec{B}\) is the magnetic field
• q = -2 \mu C = -2 \times 10^{-6} \, C
• Velocity, \(\vec{v} = (2\hat{i} + 3\hat{j}) \times 10^6 \, ms^{-1}\)
• Magnetic field, \(\vec{B} = 2\hat{j} \, T\)

First, calculate the cross product \(\vec{v} \times \vec{B}\):

\(\vec{v} \times \vec{B} = (2\hat{i} + 3\hat{j}) \times 10^6 \, ms^{-1} \times 2\hat{j} \, T\)

The cross product calculation:

\(\vec{v} \times \vec{B} = 2 \times 10^6 \times 2 (\hat{i} \times \hat{j}) + 3 \times 10^6 \times 2 (\hat{j} \times \hat{j})\)

• \(\hat{i} \times \hat{j} = \hat{k}\)
• \(\hat{j} \times \hat{j} = 0\) (since the cross product of any vector with itself is zero)

Therefore:

\(\vec{v} \times \vec{B} = 4 \times 10^6 \, \hat{k}\)

Now, calculate the magnetic force \(F\):

\(F = q \times (4 \times 10^6 \, \hat{k}) = (-2 \times 10^{-6}) \times 4 \times 10^6 \, \hat{k}\)

This simplifies to:

\(F = -8 \hat{k} \, N\)

The force is \(8 \, N\) in the negative z-direction.

Conclusion: The correct option is \(8 \, N\) in - z direction.