600
13 Electromagnetism I
f p =
Bq
2π m p
=
1.5 × 1.6 × 10 −19
2π × 1.66 × 10 −27 = 2.3 × 10
7 Hz = 23 MHz
(b) K α
1
2
(2e) 2 r 2 B 2
4m p
= K p = 13.56 MeV
f α =
(B)(2e)
2π × 4m p
=
f p
2
= 11.5 MHz
13.3 q v B = E q → v =
E
B
(1)
mv 2
r
= q v B
→ r =
mv
q B
(2)
Separation = 2r 2 − 2r 1 = 2(r 2 − r 1 )
=
2 E
q B B (m 2 − m 1 )
(3)
where we have used (1) and (2).
13.4 mv 2
r
= Bev
m =
Ber
v
=
0.2 × 1.6 × 10 −19 × 0.2
2.5 × 10 7
= 2.56 × 10
−28 kg
In terms of electron mass
m =
2.56 × 10 −28
9.1 × 10 −31 = 281 m e
Hence the particle is a pion (π – meson)
13.5 Acceleration
a =
q E
m
(1)
y =
1
2
at
2
(2)
x = vt
(3)
Combining (1), (2) and (3)
y =
q E x 2
2mv 2
(4)
which is the equation to a parabola.
13.6 r =
√
2m K
q B
=
(2 × 9.1 × 10 −31 × 5 × 10 3 × 1.6 × 10 −19 ) 1/2
1.6 × 10 −19 × 10 −2
= 0.0238 m
= 2.38 cm