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Lenz’s Law, the induced magnetic field will oppose the deflecting magnetic field.
Therefore, the deflecting magnetic field between the magnets is upward.
• Calculate the radius of the path of a proton travelling at 2.50 × 106 m/s
Fm = Fc
qvB⊥ = r= mv
r T
2 mv
qB⊥ r= T 2π r
2π m
=
qB
v
mv
r
=
qB (1.67 × 10 −27 kg)(2.50 × 106 m/s)
(1.60 × 10− 19 C)(0.863 T) r = 0.0302 m = • Calculate the final speed of a proton after it passes once between the D’s,
if its initial velocity entering the space between the D’s is 2.50 × 106 m/s.
Method I: Conservation of Energy
∆E = ∆Ek
= Vq =
vf =
= vf 1
2 mvf 2 − 1
2 mvi 2 2(Vq + 1 mvi 2 )
2
m
2[(20 000 V)(1.60 × 10− 19 C) + 1
2 (1.67 × 10− 27 kg)(2.50 × 106 m/s) 2 ] 1.67 × 10− 27 kg = 3.18 × 106 m/s Note: The incorrect solution Vq = 1
2 mvf 2 gives a speed of 1.96 × 106 m/s Method II: Dynamics and Kinematics
&
E = Fe =
Vq
d Fe V
=
q
d
&
E q = ma = ma a=
a= Vq
dm
(20 000 V)(1.60 × 10−19 C)
(5.00 × 10− 2 m)(1.67 × 10−27 kg) a = 3.832 × 1013 m/s2
vf 2 = vi 2 + 2 ad
(2.50 × 106 m/s) 2 + 2(3.832 × 1013 m/s 2 )(5.00 × 10− 2 m) vf = vf = 3.18 × 106 m/s • The speed of a particle moving with circular motion and the time it takes the
particle to complete one circular orbit are given by the formulas
v= 2π R
2π m
and T =
T
qB⊥ Beginning with force equations from the tearout Physics Data Sheet, derive the
formula for the period,
2π m
T=
qB⊥
Method I
Fm = qvB⊥ = qB⊥ = T = Method II Fc Fm mv2
r 2π R 1
m T R 2π m
qB⊥ Unit of T =
= = qvB⊥ = 2π R q B⊥
T
T 2π m
are equivalent to seconds.
qB⊥ • Show that the units of T= = 2π m
qB⊥
kg
(C)(T )
kg kg ⋅m/s2 (C) (C/s)( m) 1
1
s Unit of T = s =
= Fc
4π 2 mR
T2
4π 2 mR
T2
2π m
qB⊥...
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 Fall '09
 Quinlan
 Physics, Scientific Notation, Potential Energy, Magnetic Field, Atomic physics, .........

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