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can you please help me with these this problem and the three parts can you show it step by step
![**Electromagnetic Force on a Moving Charged Particle**
When a charged particle moves with velocity **v** through a magnetic field **B**, a force due to the magnetic field **F_B** acts on the charged particle. This occurs according to the cross-product:
\[ \mathbf{F_B} = q \mathbf{v} \times \mathbf{B} \]
where \( q \) is the charge of the particle.
### Problem Statement:
**(a)** If a particle of charge \( q = 13.4 \times 10^{-6} \) C, where the unit C is a Coulomb, moves according to the velocity vector \(\mathbf{v} = \langle 1, 5, 2 \rangle\) and the magnetic field vector is \(\mathbf{B} = \langle 4, 2, -1 \rangle\), find the force vector \(\mathbf{F_B}\) that is acting on the particle.
**(b)** What is the **magnitude** of the force on the particle?
**(c)** Sketch the right-handed system \(\{\mathbf{v}, \mathbf{B}, \mathbf{F_B}\}\) and roughly indicate the **trajectory** of the particle.
### Solution:
#### (a) Finding the Force Vector
First, compute the cross-product \(\mathbf{v} \times \mathbf{B}\):
\[
\mathbf{v} \times \mathbf{B} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 5 & 2 \\
4 & 2 & -1
\end{vmatrix}
\]
This determinant calculates to:
\[
\mathbf{v} \times \mathbf{B} = (5(-1) - 2(2))\mathbf{i} - (1(-1) - 2(4))\mathbf{j} + (1(2) - 5(4))\mathbf{k}
\]
\[
\mathbf{v} \times \mathbf{B} = (-5 - 4)\mathbf{i} - (-1 - 8)\mathbf{j} + (2 - 20)\mathbf{k}
\]
\[
\mathbf{v} \times \mathbf{B} = -9\mathbf{i} + 9](https://content.bartleby.com/qna-images/question/34cf04e3-0e6a-47ff-a4e0-f99b5389fcbb/b5dac8c9-98db-4191-be8e-c3f9b752df9f/qwvhenh_processed.png)
Transcribed Image Text:**Electromagnetic Force on a Moving Charged Particle**
When a charged particle moves with velocity **v** through a magnetic field **B**, a force due to the magnetic field **F_B** acts on the charged particle. This occurs according to the cross-product:
\[ \mathbf{F_B} = q \mathbf{v} \times \mathbf{B} \]
where \( q \) is the charge of the particle.
### Problem Statement:
**(a)** If a particle of charge \( q = 13.4 \times 10^{-6} \) C, where the unit C is a Coulomb, moves according to the velocity vector \(\mathbf{v} = \langle 1, 5, 2 \rangle\) and the magnetic field vector is \(\mathbf{B} = \langle 4, 2, -1 \rangle\), find the force vector \(\mathbf{F_B}\) that is acting on the particle.
**(b)** What is the **magnitude** of the force on the particle?
**(c)** Sketch the right-handed system \(\{\mathbf{v}, \mathbf{B}, \mathbf{F_B}\}\) and roughly indicate the **trajectory** of the particle.
### Solution:
#### (a) Finding the Force Vector
First, compute the cross-product \(\mathbf{v} \times \mathbf{B}\):
\[
\mathbf{v} \times \mathbf{B} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 5 & 2 \\
4 & 2 & -1
\end{vmatrix}
\]
This determinant calculates to:
\[
\mathbf{v} \times \mathbf{B} = (5(-1) - 2(2))\mathbf{i} - (1(-1) - 2(4))\mathbf{j} + (1(2) - 5(4))\mathbf{k}
\]
\[
\mathbf{v} \times \mathbf{B} = (-5 - 4)\mathbf{i} - (-1 - 8)\mathbf{j} + (2 - 20)\mathbf{k}
\]
\[
\mathbf{v} \times \mathbf{B} = -9\mathbf{i} + 9
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