• Support PF! Buy your school textbooks, materials and every day products Here!

Cathode Ray Tube Magnetic Question

  • #1
103
0

Homework Statement



The figure from the link below shows a cross section of a cathode ray tube. An electron in the tube initially moves horizontally in the plane of the cross section at a speed of 2.0 x 107 meters per second. The electron is deflected upward by a magnetic field that has a field strength of 6.0 x 10-4 tesla.
a. What is the direction of the magnetic field?
b. Determine the magnitude of the magnetic force acting on the electron.
c. Determine the radius of curvature of the path followed by the electron while it is in the magnetic field.

An electric field is later established in the same region as the magnetic field such that the electron now passes through the magnetic and electric fields without deflection.
d. Determine the magnitude of the electric field.
e. What is the direction of the electric field?

I found the picture at this link on page 6: harmonphys.info/magnetism%20problems.doc

Homework Equations


F = qvBsin[tex]\theta[/tex]

r = (mv)/(qB)

The Attempt at a Solution



a) It is directed out of the page by the right hand rule. (Did I do this properly?)

b) F = qvBsin[tex]\theta[/tex]
F = (1.60*10^(-19) C)(2.0*10^7 m/s)(6.0*10^(-4))sin[tex]\theta[/tex]

How do I determine [tex]\theta[/tex]?

c) r = (mv)/(qB)
r = [(9.11*10^(-31) kg)(2.0*10^7 m/s)] / [(1.60*10^(-19) C)(6.0*10^(-4) T)
r = .1898 m

d) Would the electrical and magnetic fields cancel? How do I set this up?
e) Would this be the opposite of (a)? [i.e., into the page?]
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
5

Homework Equations


F = qvBsin[tex]\theta[/tex]

r = (mv)/(qB)

The Attempt at a Solution



a) It is directed out of the page by the right hand rule. (Did I do this properly?)

b) F = qvBsin[tex]\theta[/tex]
F = (1.60*10^(-19) C)(2.0*10^7 m/s)(6.0*10^(-4))sin[tex]\theta[/tex]

How do I determine [tex]\theta[/tex]?

v and B are perpendicular. Therefore θ is 90 degrees.

c) r = (mv)/(qB)
r = [(9.11*10^(-31) kg)(2.0*10^7 m/s)] / [(1.60*10^(-19) C)(6.0*10^(-4) T)
r = .1898 m

d) Would the electrical and magnetic fields cancel? How do I set this up?
e) Would this be the opposite of (a)? [i.e., into the page?][/QUOTE]

Due to magnetic field B, the centripetal force mv^2/r bends the electron beam in the upward direction. To nullify this deflection an electric force Eq should act in the downward direction.
 
  • #3
103
0
Thank you, rl.bhat!
 

Related Threads for: Cathode Ray Tube Magnetic Question

Replies
1
Views
943
  • Last Post
Replies
1
Views
693
  • Last Post
Replies
13
Views
1K
  • Last Post
Replies
7
Views
24K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
7
Views
4K
Replies
0
Views
2K
  • Last Post
Replies
1
Views
3K
Top