24 An electron is accelerated from rest by a potential differ- ence of 350 V. It then enters a uniform magnetic field of magni- tude 200 mT with its velocity perpendicular to the field. Calculate (a) the speed of the electron and (b) the radius of its path in the magnetic field.
24 An electron is accelerated from rest by a potential differ- ence of 350 V. It then enters a uniform magnetic field of magni- tude 200 mT with its velocity perpendicular to the field. Calculate (a) the speed of the electron and (b) the radius of its path in the magnetic field.
Related questions
Question
In Problem 24's part b, assume an inbound
velocity (the value you would calculate in part (a))
of 1.11 x 107,
m/s and a magnetic field of B = 0.22
T, and solve for the radius of the circular path in
micrometers. 5 sig figs
![**Question 5**
In Problem 28.24's part b, assume an inbound velocity (the value you would calculate in part (a)) of \(1.11 \times 10^7\) m/s and a magnetic field of \(B = 0.22\) T, and solve for the radius of the circular path in micrometers. Use 5 significant figures.
[Answer Box]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe8e4ba9c-f4e2-4406-a528-e4a92bd942c6%2Feda186b8-b631-4227-aac3-3fe81669017a%2Fe2esgxd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 5**
In Problem 28.24's part b, assume an inbound velocity (the value you would calculate in part (a)) of \(1.11 \times 10^7\) m/s and a magnetic field of \(B = 0.22\) T, and solve for the radius of the circular path in micrometers. Use 5 significant figures.
[Answer Box]
![**Problem 24:**
An electron is accelerated from rest by a potential difference of 350 V. It then enters a uniform magnetic field of magnitude 200 mT with its velocity perpendicular to the field. Calculate (a) the speed of the electron and (b) the radius of its path in the magnetic field.
---
**Explanation:**
1. **Speed of the Electron:**
- Use the principle of energy conservation. The kinetic energy gained by the electron is equal to the electric potential energy lost.
- Calculate the speed using the formula:
\[
\frac{1}{2} mv^2 = e \Delta V
\]
where \( m \) is the mass of the electron, \( v \) is the speed, \( e \) is the charge of the electron, and \( \Delta V \) is the potential difference.
2. **Radius of the Path:**
- The radius of the circular path in the magnetic field is given by:
\[
r = \frac{mv}{eB}
\]
where \( B \) is the magnetic field strength.
These calculations determine how the electron’s trajectory is affected by the initial conditions and the uniform magnetic field.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe8e4ba9c-f4e2-4406-a528-e4a92bd942c6%2Feda186b8-b631-4227-aac3-3fe81669017a%2Fr7z0cl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 24:**
An electron is accelerated from rest by a potential difference of 350 V. It then enters a uniform magnetic field of magnitude 200 mT with its velocity perpendicular to the field. Calculate (a) the speed of the electron and (b) the radius of its path in the magnetic field.
---
**Explanation:**
1. **Speed of the Electron:**
- Use the principle of energy conservation. The kinetic energy gained by the electron is equal to the electric potential energy lost.
- Calculate the speed using the formula:
\[
\frac{1}{2} mv^2 = e \Delta V
\]
where \( m \) is the mass of the electron, \( v \) is the speed, \( e \) is the charge of the electron, and \( \Delta V \) is the potential difference.
2. **Radius of the Path:**
- The radius of the circular path in the magnetic field is given by:
\[
r = \frac{mv}{eB}
\]
where \( B \) is the magnetic field strength.
These calculations determine how the electron’s trajectory is affected by the initial conditions and the uniform magnetic field.
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