**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}

Given: The velocity vector is 5i⏜-4j⏜+k⏜ m/s. The magnetic field vector is i⏜+2j⏜-k⏜ T.…

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

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

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Concept explainers

Magnetic Field Of Coaxial Cable

The word coaxial means same axis. So for a long straight cable to be coaxial, it must have concentric cylindrical shaped conductor around it. Coaxial cable (popularly called ‘coax’) has various applications ranging from current conductors to radiofrequency signal transmitters. This cable has lower emission losses and provides protection from electromagnetic interference which allows signals with less power to transmit over longer distances.For example, currents produced in the pickup of a stereo turntable generally travel to the amplifier through a coaxial cable.The self-inductance of a coaxial cable is another important characteristic, because it influences the propagation of electrical signals in the cable.

Question

**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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