11. 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 x 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. (b) What is the magnitude of the force on the particle? (c) Sketch the right-handed system {v, B, FB} and roughly indicate the trajectory of the particle.
11. 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 x 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. (b) What is the magnitude of the force on the particle? (c) Sketch the right-handed system {v, B, FB} and roughly indicate the trajectory of the particle.
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![### Understanding Magnetic Force on a Charged Particle
When a charged particle moves with velocity \( \mathbf{v} \) through a magnetic field \( \mathbf{B} \), a force due to the magnetic field \( \mathbf{F_B} \) acts on the charged particle. This force is determined using the cross-product:
\[ \mathbf{F_B} = q \mathbf{v} \times \mathbf{B} \]
where \( q \) is the charge of the particle.
#### Example Problem
1. **Given:**
- Charge of the particle, \( q = 13.4 \times 10^{-6} \) C (Coulombs)
- Velocity vector, \( \mathbf{v} = \langle 1, 5, 2 \rangle \)
- Magnetic field vector, \( \mathbf{B} = \langle 4, 2, -1 \rangle \)
2. **Questions:**
(a) Find the force vector \( \mathbf{F_B} \) 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 \( \mathbf{F_B} \)**
To find \( \mathbf{F_B} \), calculate the cross-product \( \mathbf{v} \times \mathbf{B} \):
\[ \mathbf{v} = \langle 1, 5, 2 \rangle \]
\[ \mathbf{B} = \langle 4, 2, -1 \rangle \]
Using the cross-product formula for vectors in three dimensions:
\[
\mathbf{v} \times \mathbf{B} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 5 & 2 \\
4 & 2 & -1
\end{vmatrix}
\]
Solving the determinant:
\[
\mathbf{v} \times \mathbf{B} = \mathbf{i}(5 \cdot (-1) - 2 \cdot 2) - \mathbf{j}(1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F717f4d87-5a32-4cdf-855f-90bb556622e3%2Fa6068146-6b85-4403-a559-38192297774a%2F7ql3rz_processed.png&w=3840&q=75)
Transcribed Image Text:### Understanding Magnetic Force on a Charged Particle
When a charged particle moves with velocity \( \mathbf{v} \) through a magnetic field \( \mathbf{B} \), a force due to the magnetic field \( \mathbf{F_B} \) acts on the charged particle. This force is determined using the cross-product:
\[ \mathbf{F_B} = q \mathbf{v} \times \mathbf{B} \]
where \( q \) is the charge of the particle.
#### Example Problem
1. **Given:**
- Charge of the particle, \( q = 13.4 \times 10^{-6} \) C (Coulombs)
- Velocity vector, \( \mathbf{v} = \langle 1, 5, 2 \rangle \)
- Magnetic field vector, \( \mathbf{B} = \langle 4, 2, -1 \rangle \)
2. **Questions:**
(a) Find the force vector \( \mathbf{F_B} \) 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 \( \mathbf{F_B} \)**
To find \( \mathbf{F_B} \), calculate the cross-product \( \mathbf{v} \times \mathbf{B} \):
\[ \mathbf{v} = \langle 1, 5, 2 \rangle \]
\[ \mathbf{B} = \langle 4, 2, -1 \rangle \]
Using the cross-product formula for vectors in three dimensions:
\[
\mathbf{v} \times \mathbf{B} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 5 & 2 \\
4 & 2 & -1
\end{vmatrix}
\]
Solving the determinant:
\[
\mathbf{v} \times \mathbf{B} = \mathbf{i}(5 \cdot (-1) - 2 \cdot 2) - \mathbf{j}(1
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