Radius and period of charge in B field

In summary: The radius is just the length of the path multiplied by the speed. The circumference is the total distance around the path.
  • #1
velocity9
7
0

Homework Statement


A doubly charged helium atom (mass = 6.68 x 10-27 kg) is accelerated through a potential difference of 4.00x 103 V. What will be the radius of curvature of the path of the atom if it is in a uniform 0.460 T magnetic field?
Note: I hope this question is meant in advanced physics >.<
If not, I apologize. Please direct me towards the proper place.

Homework Equations


r = (mv)/(qB)

E = qV

K = ½mv^2

The Attempt at a Solution


I let
E=K and so
½mv^2 = qV
I solved for v which gave me
v = √((2*q*V)/m)
I subbed this v into
r = (mv)/(qB)

I know from the information provided that
B = 0.460 T
m = 6.68*10^-27 kg
V = 4.00*10^3
and I know q = 2*e
I also found v so there aren't any other unknowns, I subbed in...
I got 0.0199 m for r but my answer is said to be wrong. Not sure where I made a mistake. Any help is appreciated! Thank you
 
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  • #2
Welcome to PF!

Your work looks correct. Maybe you made a mistake in plugging in the numbers. What did you get for the speed v?
 
  • #3
TSny said:
Welcome to PF!

Your work looks correct. Maybe you made a mistake in plugging in the numbers. What did you get for the speed v?
I got 437740 m/s for the speed
 
  • #4
I think you might not have used the correct charge. Note that 2q = 4e.
 
  • #5
TSny said:
I think you might not have used the correct charge. Note that 2q = 4e.
gaah yes you are right! So sorry for the silly mistake. I had it written down correctly but my brain wouldn't convey the information to my fingers when plugging into the calculator. Thank you so much for your time!
 
  • #6
OK. Nice work!
 
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  • #7
TSny said:
OK. Nice work!
For the second part of the question, it asks:
What is the period of revolution for the atom in the previous question?

I'm having a little bit of trouble deciding what is related to what. I figure I can use

ω = 2π/T

the velocity found.. that is ω, right?
Not sure how else they could be connected. I tried thinking in terms of centripetal velocity but that didn't get me anywhere.
Thank you in advance!
 
  • #8
What's the length of the path that the atom follows? What's its speed around that path? How long does it take to cover that distance at that speed?
 
  • #9
gneill said:
What's the length of the path that the atom follows? What's its speed around that path? How long does it take to cover that distance at that speed?
I know the velocity since I found it in the first part. I used
velocity9 said:
v = √((2*q*V)/m)
and found v = 6.19*10^5 m/s
I think the time to cover that distance would be the period. If I could find the distance, I could probably use simple v = d/t and rearranging for t=d/v, I think the units cancel to leave you with seconds which is the unit of period. I think.
The distance I'm kind of confused about. It's not simply the radius. I would look for a circumference or something of the sort but I'm not even sure the exact shape of the path. Do I need to know if it's a circle or something like that or am I going in the wrong direction again haha
Thank you for your time!
 
  • #10
Well, you found a radius. What did you equate in order to find it? (What's it the radius of?)
 
  • #11
gneill said:
Well, you found a radius. What did you equate in order to find it? (What's it the radius of?)
Ok, so I found that mutiplying the d/v that we were discussing by 2pi works. I realize the units are convenient as we had discussed since all we are left with is seconds. I'm not entirely sure why the 2pi worked. I was searching about period and I noticed most of the periods had 2pi written next to them so I figured I would try but is it simply a matter of period having to be related with pi? Just trying to understand how my lucky guess worked. Thanks!
 
  • #12
What is the relationship between the radius and circumference of a circle?
 
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  • #13
gneill said:
What is the relationship between the radius and circumference of a circle?
Thank you so much for your patience! It all makes sense now
 

FAQ: Radius and period of charge in B field

What is the relationship between radius and period of charge in a magnetic field?

The radius and period of charge in a magnetic field are inversely proportional to each other. This means that as the radius increases, the period decreases and vice versa. This relationship is known as the Larmor frequency.

How can the radius and period of charge be calculated in a magnetic field?

The radius of a charged particle in a magnetic field can be calculated using the equation r = mv/qB, where r is the radius, m is the mass of the particle, v is its velocity, q is its charge, and B is the strength of the magnetic field. The period can then be calculated using the formula T = 2πr/v.

Can the radius and period of charge be affected by the strength of the magnetic field?

Yes, the radius and period of charge in a magnetic field are directly proportional to the strength of the magnetic field. This means that as the strength of the magnetic field increases, the radius and period also increase. This relationship is known as the cyclotron frequency.

What factors can affect the radius and period of charge in a magnetic field?

The radius and period of charge can be affected by several factors, including the strength of the magnetic field, the charge and mass of the particle, and the initial velocity of the particle. Additionally, external forces such as electric fields or gravitational fields can also influence the radius and period of charge.

How does the radius and period of charge in a magnetic field relate to circular motion?

The motion of a charged particle in a magnetic field can be described as circular motion, with the particle moving in a circular path with a constant radius and period. The strength of the magnetic field determines the radius of the circle, while the velocity of the particle determines the period of the motion.

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