An electron is moving at a speed of 4.8 × 106 m/s at an angle of 30.0° with respect to a uniform magnetic field of 9.2 × 10–4 T. What is the radius of the resulting helical path? (me = 9.11 × 10–31 kg, qe = 1.60 × 10–19 C)

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Chapter1: Units, Trigonometry. And Vectors
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An electron is moving at a speed of 4.8 × 106 m/s at an angle
of 30.0° with respect to a uniform magnetic field of 9.2 × 10–4 T.
What is the radius of the resulting helical path? (me = 9.11 × 10–31
kg, qe = 1.60 × 10–19 C)

Please use equation sheet

 

AV = -Ed
C = € °
å
R = PĀ
Δν
kelQ1||Q2l
r2
V = E, keqi
ri
F
E
Iql
v2
P = 1²R =
= IV
R
R = Ro[1+a(T – To)]
1
+
R1
1
+...
R3
AV = IR
Reg = R1 + R2 + R3+...
Req
kelQl
E =
r2
Areaπr2|
V = !
Hol
B =
2ar
F
Holil2
mv
r =
qB
F = qvB sin0
- %=
L
2nd
1
1
+
C1
1
Ceg = C1 + C2 + ….
Ceq
||
Transcribed Image Text:AV = -Ed C = € ° å R = PĀ Δν kelQ1||Q2l r2 V = E, keqi ri F E Iql v2 P = 1²R = = IV R R = Ro[1+a(T – To)] 1 + R1 1 +... R3 AV = IR Reg = R1 + R2 + R3+... Req kelQl E = r2 Areaπr2| V = ! Hol B = 2ar F Holil2 mv r = qB F = qvB sin0 - %= L 2nd 1 1 + C1 1 Ceg = C1 + C2 + …. Ceq ||
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