A proton moves perpendicular to a uniform magnetic field B at a speed of 2.10 x 10' m/s and experiences an acceleration of 1.90 x 103 m/s in the positive x-dire when its velocity is in the positive z-direction. Determine the magnitude and direction of the field. magnitude direction |---Select--- v
A proton moves perpendicular to a uniform magnetic field B at a speed of 2.10 x 10' m/s and experiences an acceleration of 1.90 x 103 m/s in the positive x-dire when its velocity is in the positive z-direction. Determine the magnitude and direction of the field. magnitude direction |---Select--- v
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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![**Problem Statement:**
A proton moves perpendicular to a uniform magnetic field \(\vec{B}\) at a speed of \(2.10 \times 10^7\) m/s and experiences an acceleration of \(1.90 \times 10^{13}\) m/s\(^2\) in the positive x-direction when its velocity is in the positive z-direction. Determine the magnitude and direction of the field.
**Solution Inputs:**
- **Magnitude:** [Input Box] T
- **Direction:** [Dropdown Menu with options]
**Explanation:**
This exercise involves calculating the magnetic field experienced by a moving proton. The problem provides the speed of the proton, its acceleration, and the directions of its velocity and acceleration. The task is to determine both the magnitude (in Tesla, T) and the direction of the magnetic field \(\vec{B}\) that the proton is moving through.
**Physics Concept:**
To solve the problem, apply the formula for the magnetic force on a moving charge:
\[ F = qvB \sin(\theta) \]
where \( F \) is the force (related to acceleration by \( F = ma \)), \( q \) is the charge of the proton, \( v \) is its velocity, \( B \) is the magnetic field, and \( \theta \) is the angle between the velocity and magnetic field vectors (90 degrees in this case as they are perpendicular).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F021773a5-30bf-4026-94da-b5b931a87d69%2Ffd858ee3-b839-4351-a878-09b3695f4911%2Fr6j6s06_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A proton moves perpendicular to a uniform magnetic field \(\vec{B}\) at a speed of \(2.10 \times 10^7\) m/s and experiences an acceleration of \(1.90 \times 10^{13}\) m/s\(^2\) in the positive x-direction when its velocity is in the positive z-direction. Determine the magnitude and direction of the field.
**Solution Inputs:**
- **Magnitude:** [Input Box] T
- **Direction:** [Dropdown Menu with options]
**Explanation:**
This exercise involves calculating the magnetic field experienced by a moving proton. The problem provides the speed of the proton, its acceleration, and the directions of its velocity and acceleration. The task is to determine both the magnitude (in Tesla, T) and the direction of the magnetic field \(\vec{B}\) that the proton is moving through.
**Physics Concept:**
To solve the problem, apply the formula for the magnetic force on a moving charge:
\[ F = qvB \sin(\theta) \]
where \( F \) is the force (related to acceleration by \( F = ma \)), \( q \) is the charge of the proton, \( v \) is its velocity, \( B \) is the magnetic field, and \( \theta \) is the angle between the velocity and magnetic field vectors (90 degrees in this case as they are perpendicular).
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