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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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**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 \
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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