An electron enters a region of space containing a uniform 2.01 x 10-5 T magnetic field. Its speed is 129 m/s and it enters perpendicularly to the field. Under these conditions, the electron undergoes circular motion. Find the radius r of the electron's path and the frequency f of the motion. r = f = m Hz

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Chapter1: Units, Trigonometry. And Vectors
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**Problem:**

An electron enters a region of space containing a uniform \(2.01 \times 10^{-5}\, \text{T}\) magnetic field. Its speed is \(129\, \text{m/s}\) and it enters perpendicularly to the field. Under these conditions, the electron undergoes circular motion. Find the radius \(r\) of the electron's path and the frequency \(f\) of the motion.

**Solution:**

To find the radius \(r\) of the electron's path:

\[ r = \]

(To be filled with the calculated answer in meters)

To find the frequency \(f\) of the motion:

\[ f = \]

(To be filled with the calculated answer in hertz)

These answers can be calculated using the formulas for motion of a charged particle in a magnetic field:

\[
r = \frac{mv}{qB}
\]

\[
f = \frac{qB}{2\pi m}
\]

Where:
- \( m \) is the mass of the electron (\( 9.11 \times 10^{-31}\, \text{kg} \))
- \( v \) is the speed of the electron (\(129\, \text{m/s}\))
- \( q \) is the charge of the electron (\( 1.60 \times 10^{-19}\, \text{C} \))
- \( B \) is the magnetic field (\(2.01 \times 10^{-5}\, \text{T}\)) 

Note: These calculations assume the motion is ideal with no external forces other than the magnetic field.
Transcribed Image Text:**Problem:** An electron enters a region of space containing a uniform \(2.01 \times 10^{-5}\, \text{T}\) magnetic field. Its speed is \(129\, \text{m/s}\) and it enters perpendicularly to the field. Under these conditions, the electron undergoes circular motion. Find the radius \(r\) of the electron's path and the frequency \(f\) of the motion. **Solution:** To find the radius \(r\) of the electron's path: \[ r = \] (To be filled with the calculated answer in meters) To find the frequency \(f\) of the motion: \[ f = \] (To be filled with the calculated answer in hertz) These answers can be calculated using the formulas for motion of a charged particle in a magnetic field: \[ r = \frac{mv}{qB} \] \[ f = \frac{qB}{2\pi m} \] Where: - \( m \) is the mass of the electron (\( 9.11 \times 10^{-31}\, \text{kg} \)) - \( v \) is the speed of the electron (\(129\, \text{m/s}\)) - \( q \) is the charge of the electron (\( 1.60 \times 10^{-19}\, \text{C} \)) - \( B \) is the magnetic field (\(2.01 \times 10^{-5}\, \text{T}\)) Note: These calculations assume the motion is ideal with no external forces other than the magnetic field.
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