Charged particle moving in circular path in a magnetic field

AI Thread Summary
A deuteron nucleus with a mass of 3.34x10^-27 kg and a charge of 1.602x10^-19 C travels in a circular path of radius 6.6 mm within a 2.1 T magnetic field. The speed of the deuteron was calculated to be approximately 6.6 x 10^5 m/s, and the time for half a rotation was determined to be about 3.14 x 10^-8 seconds. For the potential difference required to achieve this speed, calculations yielded a value of approximately 4.54 x 10^3 V. The discussion highlighted the importance of maintaining significant figures throughout calculations to avoid rounding errors. Overall, the problem was effectively solved using principles of cyclotron motion and the work-energy theorem.
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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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