**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.

Answered: Image /qna-images/answer/889f05dd-2380-446a-9572-ec58a44453a3.jpg

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

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

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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Concept explainers

Magnetic Field Of Coaxial Cable

The word coaxial means same axis. So for a long straight cable to be coaxial, it must have concentric cylindrical shaped conductor around it. Coaxial cable (popularly called ‘coax’) has various applications ranging from current conductors to radiofrequency signal transmitters. This cable has lower emission losses and provides protection from electromagnetic interference which allows signals with less power to transmit over longer distances.For example, currents produced in the pickup of a stereo turntable generally travel to the amplifier through a coaxial cable.The self-inductance of a coaxial cable is another important characteristic, because it influences the propagation of electrical signals in the cable.

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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.

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