Question

After being accelerated through a potential difference of $5.0 \mathrm{kV}$, a singly charged carbon ion $\left({ }^{12} \mathrm{C}\right)$ moves in a circle of radius $21 \mathrm{~cm}$ in the magnetic field of a mass spectrometer. What is the magnitude of the field?

          After being accelerated through a potential difference of $5.0 \mathrm{kV}$, a singly charged carbon ion $\left({ }^{12} \mathrm{C}\right)$ moves in a circle of radius $21 \mathrm{~cm}$ in the magnetic field of a mass spectrometer. What is the magnitude of the field?
        
Show more…

Added by Thomas L.

University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
AceChat toggle button
Close icon
Ace pointing down

Please give Ace some feedback

Your feedback will help us improve your experience

Thumb up icon Thumb down icon
Thanks for your feedback!
Profile picture
After being accelerated through a potential difference of $5.0 \mathrm{kV}$, a singly charged carbon ion $\left({ }^{12} \mathrm{C}\right)$ moves in a circle of radius $21 \mathrm{~cm}$ in the magnetic field of a mass spectrometer. What is the magnitude of the field?
Close icon
Play audio
Feedback
Powered by NumerAI
Kathleen Carty David Collins
Ivan Kochetkov verified

Sri K and 94 other subject Physics 101 Mechanics educators are ready to help you.

Ask a new question

*

Labs

-

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

*

Key Concepts

-
Key Concept
Premium Feature
Explore the core concept behind this problem.
Play button
Key Concept
Premium Feature
Explore the core concept behind this problem.
Your browser does not support the video tag.

*

Recommended Videos

-
the-conversion-between-atomic-mass-units-and-kilograms-is-1-mathrmu166-times-10-27-mathrmkg-after-be

The conversion between atomic mass units and kilograms is $$1 \mathrm{u}=1.66 \times 10^{-27} \mathrm{kg}$$ After being accelerated through a potential difference of $5.0 \mathrm{kV},$ a singly charged carbon ion $\left(^{12} \mathrm{C}^{+}\right)$ moves in a circle of radius $21 \mathrm{cm}$ in the magnetic ficld of a mass spectrometer. What is the magnitude of the field?

Physics

determine-the-magnitude-of-the-magnetic-field-if-an-ion-with-a-mass-twice-that-of-a-proton-and-charge-e-travels-in-a-semicircular-path-of-radius-050-m-at-a-speed-of-50-x-106-ms-mp-167e-27-kg-24411

Determine the magnitude of the magnetic field if an ion with a mass twice that of a proton and charge +e travels in a semicircular path of radius 0.50 m at a speed of 5.0 x 10^6 m/s. mp = 1.67 x 10^-27 kg and e = 1.6 x 10^-19 C

Nishant K.

determine-the-magnitude-of-the-magnetic-field-if-an-ion-with-a-mass-twice-that-of-a-proton-and-charge-e-travels-in-a-semicircular-path-of-radius-050-m-at-a-speed-of-50-x-106-ms-mp-167e-27-kg-24411

Determine the magnitude of the magnetic field if an ion with a mass twice that of a proton and charge +e travels in a semicircular path of radius 0.50 m at a speed of 5.0 x 10^6 m/s. mp = 1.67 x 10^-27 kg and e = 1.6 x 10^-19 C

Nishant K.


*

Recommended Textbooks

-
University Physics with Modern Physics

University Physics with Modern Physics

Hugh D. Young 14th Edition
achievement 1,079 solutions
Physics: Principles with Applications

Physics: Principles with Applications

Douglas C. Giancoli 7th Edition
achievement 1,588 solutions
Fundamentals of Physics

Fundamentals of Physics

David Halliday, Robert Resnick , Jearl Walker 10th Edition
achievement 1,310 solutions

*

Transcript

-
00:02 So what happens here is actually potential difference accelerating the charge particle is given as v is equal to 5 kilo volt or we can say it as 5 ,000 volts.
00:13 Okay, now as it is singly ionized carbon ion, so q will be how much one electron is having the charge of 1 .6 multiplied by 10 to the power minus 90.
00:25 Okay, cool up.
00:27 This is the charge.
00:28 Okay, now our radius is given as 21.
00:32 Centimeter or we can write it as 0 .21 meter.
00:36 So this is the data which is given.
00:38 So using equation so r is equal to under root 2 product m into charge and voltage divided by b into q.
00:51 We need to calculate b that is magnetic field.
00:53 B we do not know we need to calculate b.
00:56 So how will calculate so b will be is equal to let us calculate r then we'll calculate b.
01:02 Okay so so r is already given in the question.
01:04 What is the r? r is 0 .21.
01:07 Just now we calculated.
01:08 We put the value of r here...
Need help? Use Ace
Ace is your personal tutor. It breaks down any question with clear steps so you can learn.
Start Using Ace
Ace is your personal tutor for learning
Step-by-step explanations
Instant summaries
Summarize YouTube videos
Understand textbook images or PDFs
Study tools like quizzes and flashcards
Listen to your notes as a podcast
Continue solving this problem
Create a free account to:
  • View full step-by-step solution
  • Ask follow-up questions with Ace AI
  • Save progress and study later
Continue Free
Numerade

Get step-by-step video solution
from top educators

Continue with Clever
or



By creating an account, you agree to the Terms of Service and Privacy Policy
Already have an account? Log In

A free answer
just for you

Watch the video solution with this free unlock.

Numerade

Log in to watch this video
...and 100,000,000 more!


EMAIL

PASSWORD

OR
Continue with Clever