Charged particle moving in circular path in a magnetic field

In summary: The charge is squared in the formula for kinetic energy, but not in the formula for potential difference. So if you want to solve for potential difference, you need to take the square root of the kinetic energy instead of the factor 1/2. But it's not a good idea to work with formulae like this. It's better to work with conservation of energy, which is much more natural.
  • #1
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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  • #2
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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  • #3
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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  • #4
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?
 
  • #5
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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  • #6
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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  • #7
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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  • #8
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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Related to Charged particle moving in circular path in a magnetic field

What is a charged particle moving in circular path in a magnetic field?

A charged particle moving in circular path in a magnetic field refers to a scenario where a charged particle, such as an electron, is moving along a circular path while being subjected to a magnetic field. The magnetic field exerts a force on the charged particle, causing it to move in a circular motion.

What is the direction of the force on a charged particle moving in circular path in a magnetic field?

The direction of the force on a charged particle moving in circular path in a magnetic field is perpendicular to both the direction of the particle's velocity and the direction of the magnetic field. This is known as the right-hand rule.

What is the equation for calculating the magnitude of the force on a charged particle in a magnetic field?

The equation for calculating the magnitude of the force on a charged particle in a magnetic field is F = qvB, where F is the force, q is the charge of the particle, v is its velocity, and B is the strength of the magnetic field.

What factors affect the magnitude of the force on a charged particle in a magnetic field?

The magnitude of the force on a charged particle in a magnetic field is affected by the strength of the magnetic field, the charge of the particle, and the speed of the particle. A stronger magnetic field, a higher charge, or a higher speed will result in a greater force on the particle.

What is the relationship between the radius of the circular path and the velocity of the charged particle in a magnetic field?

The radius of the circular path and the velocity of the charged particle in a magnetic field are inversely proportional. This means that as the velocity of the particle increases, the radius of the circular path decreases, and vice versa.

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