Magnetic Force on a Charged Particle
Question: A charged particle of charge \( q = -2\,\mu\text{C} \) is in a magnetic field of \( B = 2\,\text{T} \) that acts in the \( y \)-direction. The particle’s velocity is \[ \vec{v} = \left(2\hat{i} + 3\hat{j}\right) \times 10^6\,\text{m/s}. \] Find the magnetic force acting on the particle.
Detailed Step-by-Step Explanation
Step 1: Write Down the Given Quantities
The charge on the particle is
\[
q = -2 \times 10^{-6}\,\text{C}.
\]
The magnetic field is
\[
\vec{B} = 2\,\hat{j}\,\text{T},
\]
and the velocity of the particle is
\[
\vec{v} = \left(2\hat{i} + 3\hat{j}\right) \times 10^6\,\text{m/s}.
\]
Step 2: Use the Lorentz Force Law
The magnetic force on a moving charge is given by:
\[
\vec{F} = q \, (\vec{v} \times \vec{B}).
\]
Step 3: Compute the Cross Product \( \vec{v} \times \vec{B} \)
With
\[
\vec{v} = \left(2\hat{i} + 3\hat{j}\right) \times 10^6\,\text{m/s} \quad \text{and} \quad \vec{B} = 2\hat{j}\,\text{T},
\]
we have:
\[ \vec{v} \times \vec{B} = \left(2\hat{i} \times 2\hat{j}\right) \times 10^6 + \left(3\hat{j} \times 2\hat{j}\right) \times 10^6. \]
Note that \( \hat{j} \times \hat{j} = 0 \) and \( \hat{i} \times \hat{j} = \hat{k} \), so:
\[ \vec{v} \times \vec{B} = 4 \times 10^6\,\hat{k}\,\text{m/s}. \]
Step 4: Calculate the Magnetic Force
Now substitute into the Lorentz force formula:
\[
\vec{F} = q \, (\vec{v} \times \vec{B}) = \left(-2 \times 10^{-6}\,\text{C}\right) \times \left(4 \times 10^6\,\hat{k}\,\text{m/s}\right).
\]
Simplify the multiplication:
\[
\vec{F} = -8\,\hat{k}\,\text{N}.
\]
The negative sign indicates that the force is directed in the negative \( z \)-direction. Thus, the magnitude of the force is \( 8\,\text{N} \) directed along the \( z \)-axis.
Final Answer: The magnetic force acting on the particle is \( 8\,\text{N} \) in the \( z \)-direction, which corresponds to option (c).
.png)
0 Comments