Charged particle moving in circular path in a magnetic field

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Homework Help Overview

The discussion revolves around a deuteron nucleus moving in a circular path within a magnetic field, focusing on determining its speed, the time for half a rotation, and the potential difference required for acceleration. The subject area includes concepts of cyclotron motion and the work-energy theorem.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between radius, speed, and magnetic field strength, with one participant attempting to derive the speed using the formula for circular motion. Questions arise regarding the appropriate equations for calculating time and potential difference.

Discussion Status

Some participants have provided guidance on calculating the time for half a rotation based on known values, while others have discussed the application of the work-energy theorem for potential difference. Multiple interpretations of the calculations are being explored, with no explicit consensus reached on the final values.

Contextual Notes

Participants note the importance of significant figures in calculations to avoid rounding errors, and there are indications of arithmetic discrepancies in the potential difference calculations. The original poster expresses uncertainty in the approach to part B and seeks clarification.

cookiemnstr510510
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Homework Statement


A deuteron nucleus (consisting of one proton and one neutron) has a mass of 3.34x10-27kg and a charge of 1.602x10-19C. The deuteron nucleus travels in a circular path of radius, 6.6mm, in a magnetic field with magnitude of 2.1T.
A) Find the speed of the deuteron nucleus
B) Find the time required to make one-half of a complete rotation
C) Through what potential difference would the deuteron nucleus have to be accelerated in order to acquire this speed?

Homework Equations


F=qvB=mar=(mv2)/r
r=(mv)/(qB)

The Attempt at a Solution


I believe this situation is considered cyclotron motion.

A)
##r=\frac{mv}{qB}##→##v=\frac{rqB}{m}##= ##\frac{(6.6*10^-3m)(1.602*10^-19C)(2.1T)}{3.34*10^-27kg}##=6.6*105##\frac{m}{s}##
B)I am stuck on B, which equations should I be using?
C) Ill get to c after I figure out B!

Any help will be appreciated!
 
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It's moving in a circular path. You know the radius, so you know the circumference of the path. You know the speed. So calculating how long it takes to go half way around should be easy. You don't need a formula. Just think about it.
 
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For c) Use the work-energy theorem together with the fact that the work of the electric field with potential difference V is simply W=Vq (I guess you have seen this equation as V=W/q) , where q the charge of the particle.
 
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Okay I've figured out B:

1 rotation=2πr; 0.5 rotation=πr

Velocity=##\frac{distance}{time}##, two of these variables are known.

Time=##\frac{d}{v}##=##\frac{(π)(6.6*10^-3m)}{6.6*10^-3m/s}##=3.14*10-8s

For C:
W=ΔK=Kf-Ki→Vq=##\frac{1}{2}##mv2→V=##\frac{(1/2)mv^2}{q}##=##\frac{(1/2)(3.34*10^-27kg)(6.6*10^5m/s)^2}{1.602*10^-19C}##=2.3*10^3 V?
 
For the c) I get V=4.54 x 10^3 V. I think you just made a mistake in the arithmetic operations.
 
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You should keep more significant figures in intermediate results. That way rounding error won't creep into your significant figures as you proceed through subsequent calculations with the values. Only round at the end to present results.
 
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You've done B correctly and I think this is the right answer. But you have the speed written as 6.6E-3 m/s in the denominator and it should be 6.6E5 m/s.
 
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Delta² said:
For the c) I get V=4.54 x 10^3 V. I think you just made a mistake in the arithmetic operations.
I'd guess the factor 1/2 was applied twice.
 
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