When a charged particle moves with velocity v through a magnetic field B, a force due to the magnetic field FB acts on the charged particle. This occurs according to the cross-product: FB = qv × B where q is the charge of the particle. (a) If a particle of charge q = 13.4 x 10-6C, where the unit C is a Coulomb, moves according to the velocity vector v = (1,5, 2) and the magnetic field vector is B = (4, 2, -1), find the force vector FB that is acting on the particle. 5) What is the magnitude of the force on the particle? =) Sketch the right-handed system {v, B, FB} and roughly indicate the trajectory of the particle.

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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
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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