A proton moves with a velocity of v = (Sî – 4j + k) m/s in a region in which the magnetic field is B = (î + 2j – k) T. What is the magnitude of the magnetic force this particle experiences? %3D %3D

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Chapter1: Units, Trigonometry. And Vectors
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**Problem Statement:**

A proton moves with a velocity of \(\vec{v} = (5\hat{i} - 4\hat{j} + \hat{k}) \, \text{m/s}\) in a region in which the magnetic field is \(\vec{B} = (\hat{i} + 2\hat{j} - \hat{k}) \, \text{T}\). What is the magnitude of the magnetic force this particle experiences?

**Solution Explanation:**

The magnetic force \(\vec{F}\) experienced by a charged particle moving in a magnetic field is given by the equation:

\[
\vec{F} = q (\vec{v} \times \vec{B})
\]

where \(q\) is the charge of the proton, \(\vec{v}\) is the velocity vector, and \(\vec{B}\) is the magnetic field vector.

**Cross Product Calculation:**

To find \(\vec{v} \times \vec{B}\), use the determinant method:

\[
\vec{v} \times \vec{B} = \begin{vmatrix} 
\hat{i} & \hat{j} & \hat{k} \\
5 & -4 & 1 \\
1 & 2 & -1 \\
\end{vmatrix}
\]

Calculate each component of the cross product:

- \(\hat{i}\) component: \((-4)(-1) - (1)(2) = 4 - 2 = 2\)
- \(\hat{j}\) component: \(-( (5)(-1) - (1)(1) ) = -(-5 - 1) = 6\)
- \(\hat{k}\) component: \((5)(2) - (-4)(1) = 10 + 4 = 14\)

Thus, \(\vec{v} \times \vec{B} = (2\hat{i} + 6\hat{j} + 14\hat{k})\).

**Magnitude of the Magnetic Force:**

The magnitude \(|\vec{F}|\) is given by:

\[
|\vec{F}| = q |\vec{v} \times \vec{B}|
\]

The charge of a proton \(q = 1.6 \times 10^{-19}
Transcribed Image Text:**Problem Statement:** A proton moves with a velocity of \(\vec{v} = (5\hat{i} - 4\hat{j} + \hat{k}) \, \text{m/s}\) in a region in which the magnetic field is \(\vec{B} = (\hat{i} + 2\hat{j} - \hat{k}) \, \text{T}\). What is the magnitude of the magnetic force this particle experiences? **Solution Explanation:** The magnetic force \(\vec{F}\) experienced by a charged particle moving in a magnetic field is given by the equation: \[ \vec{F} = q (\vec{v} \times \vec{B}) \] where \(q\) is the charge of the proton, \(\vec{v}\) is the velocity vector, and \(\vec{B}\) is the magnetic field vector. **Cross Product Calculation:** To find \(\vec{v} \times \vec{B}\), use the determinant method: \[ \vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & -4 & 1 \\ 1 & 2 & -1 \\ \end{vmatrix} \] Calculate each component of the cross product: - \(\hat{i}\) component: \((-4)(-1) - (1)(2) = 4 - 2 = 2\) - \(\hat{j}\) component: \(-( (5)(-1) - (1)(1) ) = -(-5 - 1) = 6\) - \(\hat{k}\) component: \((5)(2) - (-4)(1) = 10 + 4 = 14\) Thus, \(\vec{v} \times \vec{B} = (2\hat{i} + 6\hat{j} + 14\hat{k})\). **Magnitude of the Magnetic Force:** The magnitude \(|\vec{F}|\) is given by: \[ |\vec{F}| = q |\vec{v} \times \vec{B}| \] The charge of a proton \(q = 1.6 \times 10^{-19}
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