Electromagnetic Force

1. Why does the image become distorted?
2. Why is the image covered with colored bands of red, green, and blue?
3. Why does the color stay screwed up after the magnet is removed?
4. Why does the color return to normal after the television has been turned off and back on?
5. Why doesn't this effect occur with LED, LCD, and plasma displays?

1. The beam enters region a. with negligible initial velocity and is accelerated across a potential difference of 10,000 V in a distance of 10 cm.

   

   ΔV =	10,000 V = 104 V
   d =	10 cm = 0.10 m = 10−1 m
   v0 =	0 m/s
   q =	e = 1.60 × 10−19 C
   m =	9.11 × 10−31 kg

   1. the magnitude of the electric field E_a is…

      E =  ∆V d

      Ea =  104 V 10−1 m

      Ea = 105 V/m = 105 N/C

   2. the electric force on an electron in this region is…

      FE = qE  ⇐  E =  FE q

      FE = (1.60 × 10−19 C)(105 N/C)

      FE = 1.60 × 10−14 N toward the right side of the diagram

   3. the final speed of the electrons when they exit this region is…

      either kinematics

      v2 = v02 + 2ad

      v = √(2ad)

      Newton's second law of motion

      a =  F m

      ⇐

      F = ma

      and algebra

      v = √	2FEd
      	m

      or work-energy theorem

      W =	∆E
      Fd =	½mv2

      and algebra

      v = √	2FEd
      	m

      same difference

      v = √	2(1.60 × 10−14 N)(10−1 m)
      	(9.11 × 10−31 kg)

      v = 5.93 × 10⁷ m/s

2. The beam then passes into region b. where there are both electric and magnetic fields. The electric field strength in region b. has the same magnitude as region a. but points downward. The magnetic field points inward and is adjusted so that the magnetic and electric forces on the electron beam cancel.

   

   Eb =	Ea = 104 V/m
   FB =	FE = 1.60 × 10−14 N
   v =	5.93 × 107 m/s
   q =	e = 1.60 × 10−19 C
   m =	9.11 × 10−31 kg

   4. the magnitude of the magnetic field in this region

      B =  FB qv  ⇐  FB = qvB sin θ

      Bb =  1.60 × 10−14 N (1.60 × 10−19 C)(5.93 × 107 m/s)

      Bb = 1.69 × 10−3 T

      or do it from first principles

      FB =	FE
      qvB =	qE
      Bb =	E v
      Bb =	105 V/m 5.93 × 107 m/s
      Bb =	1.69 × 10−3 T

3. The beam finally enters region c., where there is a magnetic field but no electric field. The electrons follow a semicircular path with radius 7.5 cm.

   

   v =	5.93 × 107 m/s
   q =	e = 1.60 × 10−19 C
   m =	9.11 × 10−31 kg

   5. the magnitude and direction of the magnetic field in this region

      FB =	FC
      qvB =	mv2 r
      Bc =	mv qr
      Bc =	(9.11 × 10−31 kg)(5.93 × 107 m/s) (1.60 × 10−19 C)(0.075 m)
      Bc =	4.50 × 10−3 T out of the diagram

      This information could be added to the diagram by drawing dots to indicate field lines coming out. But how densely should these dots be packed? Let's compare the magnitude of the magnetic field in region c to region b.

      Bc	=	4.499… × 10−3 T	=	2⅔ or 83 exactly
      Bb		1.687… × 10−3 T

      Well that's interesting. Looking for a challenge? See if you can show why this is true without using a lot of algebra. (Some algebra is needed, of course, but less would be better.)

Answer it.

Answer it.