https://file.aiquickdraw.com/mathbot/file/1729863073891ap77qvp3.jpg

Let's solve this problem step by step:

### Given:
- Magnetic field (\(B\)) = 4.8 mT = \(4.8 \times 10^{-3}\) T.
- Magnetic force (\(F\)) = \(3.3 \times 10^{-15}\) N.
- The particle can be either a proton or an electron. We'll determine which one it is.

### Relevant Formulas:
1. **Magnetic Force**: \( F = qvB \)
   - \( q \): charge of the particle (\(1.6 \times 10^{-19}\) C for both protons and electrons).
   - \( v \): speed of the particle.
2. **Radius of the Circular Path**: \( r = \frac{mv}{qB} \)
   - \( m \): mass of the particle.
   - For a proton, \( m_p = 1.67 \times 10^{-27} \) kg.
   - For an electron, \( m_e = 9.11 \times 10^{-31} \) kg.
3. **Period of Motion**: \( T = \frac{2\pi r}{v} \) or \( T = \frac{2\pi m}{qB} \).

### Step-by-Step Solution:

#### Step 1: Determine the Type of Particle
We use the formula \( F = qvB \) to find \( v \):
\[
v = \frac{F}{qB}
\]

Substituting the values:
- \( q = 1.6 \times 10^{-19} \) C
- \( F = 3.3 \times 10^{-15} \) N
- \( B = 4.8 \times 10^{-3} \) T

\[
v = \frac{3.3 \times 10^{-15}}{1.6 \times 10^{-19} \times 4.8 \times 10^{-3}}
\]
\[
v \approx 4.30 \times 10^6 \, \text{m/s}
\]

Since this speed is within the typical range for electrons in such fields, we choose the particle to be an electron.

#### Step 2: Calculate the Radius (\( r \))
Using the formula \( r = \frac{mv}{qB} \):
- \( m = 9.11 \times 10^{-31} \) kg (mass of an electron)

\[
r = \frac{9.11 \times 10^{-31} \times 4.30 \times 10^6}{1.6 \times 10^{-19} \times 4.8 \times 10^{-3}}
\]
\[
r \approx 5.1 \times 10^{-2} \, \text{m} = 5.1 \, \text{cm}
\]

#### Step 3: Calculate the Period (\( T \))
Using the formula \( T = \frac{2\pi m}{qB} \):
\[
T = \frac{2\pi \times 9.11 \times 10^{-31}}{1.6 \times 10^{-19} \times 4.8 \times 10^{-3}}
\]
\[
T \approx 7.5 \times 10^{-11} \, \text{s}
\]

### Final Answers:
- (a) The particle's speed: \( 4.30 \times 10^6 \, \text{m/s} \)
- (b) The radius of the circle: \( 5.1 \, \text{cm} \)
- (c) The period of the motion: \( 7.5 \times 10^{-11} \, \text{s} \)

Would you like a detailed explanation of any part of this process or more information on the concepts used?

### Related Questions:
1. How does the magnetic force affect the circular motion of a charged particle?
2. What is the significance of choosing a proton or an electron in this problem?
3. How would the results change if the magnetic field strength increased?
4. Why do electrons typically move faster in a magnetic field than protons?
5. How can the radius of the circular motion be experimentally determined?

### Tip:
Remember that in a uniform magnetic field, charged particles move in circular paths due to the centripetal force provided by the magnetic force. The radius and speed depend on the charge, mass, and strength of the field.

In the figure, a particle moves along a circle in a region of uniform magnetic field of magnitude B = 4.8 mT. The particle is either a proton or an electron (you must decide which). It experiences a magnetic force of magnitude 3.3 × 10^(-15) N. What are (a) the particle's speed, (b) the radius of the circle, and (c) the period of the motion?

Learn how to calculate the speed, radius, and period of an electron moving in a magnetic field of 4.8 mT. This detailed solution uses the Lorentz force formula and explains step-by-step how the magnetic force influences the particle's circular motion. Includes calculations and insights for understanding the relationship between magnetic field strength, charge, and velocity.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation