# Charged particle moving in circular path in a magnetic field

1. 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?

2. Homework Equations
F=qvB=mar=(mv2)/r
r=(mv)/(qB)

3. 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!

Related Introductory Physics Homework Help News on Phys.org

#### phyzguy

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.

#### Delta2

Homework Helper
Gold Member
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.

Okay Ive 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?

#### Delta2

Homework Helper
Gold Member
For the c) I get V=4.54 x 10^3 V. I think you just made a mistake in the arithmetic operations.

#### gneill

Mentor
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.

#### phyzguy

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.

#### haruspex

Homework Helper
Gold Member
2018 Award
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.

"Charged particle moving in circular path in a magnetic field"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving