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? =) Sketch the right-handed system {v, B, FB} and roughly indicate the trajectory of the particle.
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? =) Sketch the right-handed system {v, B, FB} and roughly indicate the trajectory of the particle.
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Can you help me with these three. parts and can you show me how to do it step by step solving these three. problems. so I CAN UNDERStand it better
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![**Understanding the Force on a Charged Particle in a Magnetic Field**
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 defined as:
\[ \mathbf{F_B} = q\mathbf{v} \times \mathbf{B} \]
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 { \(\mathbf{v, B, F_B}\) } and roughly indicate the trajectory of the particle.
### Solution:
**(a)** To find the force vector \( \mathbf{F_B} \) acting on the particle, we need to calculate the cross product \( \mathbf{v} \times \mathbf{B} \):
\[ \mathbf{v} = \langle 1, 5, 2 \rangle \]
\[ \mathbf{B} = \langle 4, 2, -1 \rangle \]
The cross product \( \mathbf{v} \times \mathbf{B} \) is calculated as follows:
\[
\mathbf{v} \times \mathbf{B} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 5 & 2 \\
4 & 2 & -1
\end{vmatrix}
\]
Expanding the determinant, we get:
\[
\mathbf{v} \times \mathbf{B} = \mathbf{i} (5 \cdot -1 - 2 \cdot 2) - \mathbf{j} (1 \cdot -1 - 2 \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0ee3a518-8279-4c7a-b55b-ef9ad8090828%2Fded37a63-d18d-4aa9-b294-3ed12f07b665%2F5jiuzan_processed.png&w=3840&q=75)
Transcribed Image Text:**Understanding the Force on a Charged Particle in a Magnetic Field**
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 defined as:
\[ \mathbf{F_B} = q\mathbf{v} \times \mathbf{B} \]
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 { \(\mathbf{v, B, F_B}\) } and roughly indicate the trajectory of the particle.
### Solution:
**(a)** To find the force vector \( \mathbf{F_B} \) acting on the particle, we need to calculate the cross product \( \mathbf{v} \times \mathbf{B} \):
\[ \mathbf{v} = \langle 1, 5, 2 \rangle \]
\[ \mathbf{B} = \langle 4, 2, -1 \rangle \]
The cross product \( \mathbf{v} \times \mathbf{B} \) is calculated as follows:
\[
\mathbf{v} \times \mathbf{B} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 5 & 2 \\
4 & 2 & -1
\end{vmatrix}
\]
Expanding the determinant, we get:
\[
\mathbf{v} \times \mathbf{B} = \mathbf{i} (5 \cdot -1 - 2 \cdot 2) - \mathbf{j} (1 \cdot -1 - 2 \
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