3. A proton moving eastward with a velocity of 5.0 km/s enters a magnetic field of 0.20 T pointing northward. What are the magnitude and direction of the force that the magnetic field exerts on the proton? (qe=+1.60 × 10-19 C)
3. A proton moving eastward with a velocity of 5.0 km/s enters a magnetic field of 0.20 T pointing northward. What are the magnitude and direction of the force that the magnetic field exerts on the proton? (qe=+1.60 × 10-19 C)
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 moving eastward with a velocity of 5.0 km/s enters a magnetic field of 0.20 T pointing northward. What are the **magnitude and direction** of the force that the magnetic field exerts on the proton?
*(qₑ = +1.60 × 10⁻¹⁹ C)*
---
**Explanation:**
To solve this problem, we use the formula for the magnetic force on a moving charge:
\[ F = qvB \sin(\theta) \]
Where:
- \( F \) is the force,
- \( q \) is the charge of the proton (\( 1.60 \times 10^{-19} \) C),
- \( v \) is the velocity of the proton (5.0 km/s or 5000 m/s),
- \( B \) is the magnetic field strength (0.20 T),
- \( \theta \) is the angle between the velocity and the magnetic field.
Since the velocity is eastward and the magnetic field is northward, the angle \( \theta \) between them is 90 degrees. Thus, \( \sin(90°) = 1 \).
Substitute the values into the formula:
\[ F = (1.60 \times 10^{-19} \, \text{C})(5000 \, \text{m/s})(0.20 \, \text{T}) \]
\[ F = 1.60 \times 10^{-19} \, \text{C} \times 1000 \, \text{T} \cdot \text{m/s} \]
\[ F = 3.2 \times 10^{-16} \, \text{N} \]
**Direction:**
The direction of the force is determined using the right-hand rule. Point your fingers in the direction of the velocity (east), and curl them toward the magnetic field direction (north). Your thumb will point in the direction of the force, which in this case is upwards or out of the page.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0d0fca7b-87e2-42be-9458-260269753889%2Fd52019de-8001-4366-9408-c6873381a5ad%2Fzpi1t2jk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A proton moving eastward with a velocity of 5.0 km/s enters a magnetic field of 0.20 T pointing northward. What are the **magnitude and direction** of the force that the magnetic field exerts on the proton?
*(qₑ = +1.60 × 10⁻¹⁹ C)*
---
**Explanation:**
To solve this problem, we use the formula for the magnetic force on a moving charge:
\[ F = qvB \sin(\theta) \]
Where:
- \( F \) is the force,
- \( q \) is the charge of the proton (\( 1.60 \times 10^{-19} \) C),
- \( v \) is the velocity of the proton (5.0 km/s or 5000 m/s),
- \( B \) is the magnetic field strength (0.20 T),
- \( \theta \) is the angle between the velocity and the magnetic field.
Since the velocity is eastward and the magnetic field is northward, the angle \( \theta \) between them is 90 degrees. Thus, \( \sin(90°) = 1 \).
Substitute the values into the formula:
\[ F = (1.60 \times 10^{-19} \, \text{C})(5000 \, \text{m/s})(0.20 \, \text{T}) \]
\[ F = 1.60 \times 10^{-19} \, \text{C} \times 1000 \, \text{T} \cdot \text{m/s} \]
\[ F = 3.2 \times 10^{-16} \, \text{N} \]
**Direction:**
The direction of the force is determined using the right-hand rule. Point your fingers in the direction of the velocity (east), and curl them toward the magnetic field direction (north). Your thumb will point in the direction of the force, which in this case is upwards or out of the page.
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