Radius of ions path in a field

F=qvB=ma) to solve for v. once you have v, then you can plug it into the equation for r and solve for the radius of the ion's path in the magnetic field.
  • #1
rinarez7
27
0
1. A singly charged positive ion has a mass = 1.08e-26 kg
After being accelerated through a potential difference of 240 V the ion enters a magnetic field of 0.595 T in a direction perpendicular to the field.
The charge on the ion is 1.602 e-19 C.
Find the radius of the ion's path in the field.
Answer in units of cm.



2. m= 1.08e-26 kg
potential difference= 240 V
B= 0.595 T
q= 1.602e-19
sin theta= 1 (enters in direction perpendicular to the field)
radius= mv/ qB




3. I am confused on exactly how to calculate velocity and the relevance of the potential difference. But, I tried:
F=qvBsin theta= ma
F=qvB= ma
1.602e-19(v) (0.595T)= 1.08e-26(-9.81m/s^2)
v= 1.112e-6 m/s
I then plugged this into r= mv/ qB
r= 1.08e-26kg (1.112e-6 m/s )/ 1.602e-19 (0.595T) and calculated r= 1.8997 e-11 cm, but this incorrect. What am I missing?
 
Physics news on Phys.org
  • #2
Hi rinarez7

rinarez7 said:
1. A singly charged positive ion has a mass = 1.08e-26 kg
After being accelerated through a potential difference of 240 V the ion enters a magnetic field of 0.595 T in a direction perpendicular to the field.
The charge on the ion is 1.602 e-19 C.
Find the radius of the ion's path in the field.
Answer in units of cm.



2. m= 1.08e-26 kg
potential difference= 240 V
B= 0.595 T
q= 1.602e-19
sin theta= 1 (enters in direction perpendicular to the field)
radius= mv/ qB

3. I am confused on exactly how to calculate velocity and the relevance of the potential difference. But, I tried:
F=qvBsin theta= ma
F=qvB= ma
1.602e-19(v) (0.595T)= 1.08e-26(-9.81m/s^2)

the acceleration to use in this problem is not gravity! in fact you can likely neglect gravity... check at the end that your calculated accerlation is much larger than g.

the ion will undergo circular motion due to the magnetic field, and i think you have used this to find r, however v is unknown as you don't yet know a...

rinarez7 said:
v= 1.112e-6 m/s
I then plugged this into r= mv/ qB
r= 1.08e-26kg (1.112e-6 m/s )/ 1.602e-19 (0.595T) and calculated r= 1.8997 e-11 cm, but this incorrect. What am I missing?

to find v, think about the energy of the ion due to being accelerated by the potential difference and how it relates to velocity
 

1. What is the radius of an ion's path in a uniform electric field?

The radius of an ion's path in a uniform electric field is determined by the ion's charge, mass, and the strength of the electric field. It can be calculated using the formula: r = (m*v)/(q*B), where m is the mass of the ion, v is its velocity, q is its charge, and B is the strength of the magnetic field.

2. How does the radius of an ion's path change with increasing electric field strength?

As the electric field strength increases, the radius of an ion's path will also increase. This is because the force on the ion will be greater, causing it to curve more in its path.

3. What is the effect of ion mass on the radius of its path in a field?

The mass of the ion has a direct effect on the radius of its path in a field. A heavier ion will have a larger radius, while a lighter ion will have a smaller radius. This is due to the fact that the force on the ion is proportional to its mass.

4. How does the charge of an ion affect the radius of its path in a field?

The charge of an ion also has a direct effect on the radius of its path in a field. A higher charge will result in a larger radius, while a lower charge will result in a smaller radius. This is because the force on the ion is directly proportional to its charge.

5. Can the radius of an ion's path in a field be altered?

Yes, the radius of an ion's path in a field can be altered by changing the strength of the electric field or by altering the ion's charge or mass. Additionally, if a magnetic field is present, it can also affect the radius of the ion's path.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
140
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
6K
  • Introductory Physics Homework Help
Replies
1
Views
996
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
931
  • Introductory Physics Homework Help
Replies
3
Views
995
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Advanced Physics Homework Help
Replies
4
Views
3K
Back
Top