Problem#1
An electron is shot into one end of a solenoid. As it enters the uniform magnetic field within the solenoid, its speed is 800 m/s and its velocity vector makes an angle of 30° with the central axis of the solenoid. The solenoid carries 4.0 A and has 8000 turns along its length. How many revolutions does the electron make along its helical path within the solenoid by the time it emerges from the solenoid’s opposite end? (In a real solenoid, where the field is not uniform at the two ends, the number of revolutions would be slightly less than the answer here.)Answer:
Known:
Speed, v = 800 m/s
carries current, i = 4.0 A
number of turns, N = 8000
find the magnetic field;
B = μ0iN/L = (4π x 10─7 T.m/A)(4.0 A)(8000)/L
B = 0.0402/L
The radius of the electron path:
R = mvsin 300/eB = (9.11 x 10─31 kg)(800 m/s)(0.5)/(1.6 x 10─19 C x 0.0402/L)
R = 5.66 x 10─8L
Period of the electron movement:
T = 2πR/(v sin 300) = 2π x 5.66 x 10─8L/(800 m/s x 0.5) = 8.895 x 10─10 L
Horizontal distance:
d = (v cos 300)T = 800 m/s x ½ √3 x 8.895 x 10─10 L = 6.1629 x 10─7L
Therefore:
L/d = number of revolutions (N) = 1/(6.1629 x 10─7)
N = 1.62 x 106 turn
Problem#2
A long solenoid with 10.0 turns/cm and a radius of 7.00 cm carries a current of 20.0 mA. A current of 6.00 A exists in a straight conductor located along the central axis of the solenoid. (a) At what radial distance from the axis will the direction of the resulting magnetic field be at 45.0° to the axial direction? (b) What is the magnitude of the magnetic field there?
Answer:
Known:
number of turns, n = 10 turn/cm = 1000 turn/m
radius, r = 7.00 cm = 0.07 m
(a) the magnetic field due to a solenoid is given by:
BS = μ0nIS
And the magnetic field due to a wire is given by:
Bw = μ0Iw/2πr
The angle of the net magnetic field from the axial direction is given by:
tan θ = Bw/Bs = [μ0Iw]/[2πr(μ0nIS)]
r = Iw/(2πnIstan θ)
r = (6.00 A)/(2π x 1000/m x 0.02 A x tan 450) = 0.0477 m = 4.77 m
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| Fig.1 |
(b) the magnetic field strength is given by:
B = (Bw2 + Bs2)1/2
Because
Bw/Bs = tan 450 = 1
Bw = Bs
So that
B = (Bs2 + Bs2)1/2 = Bs√2
B = μ0nIS√2 = (4π x 10─7 T.m/A)(1000/m x 0.020 A)√2 = 3.55 x 10─5 T
Problem#3
It is desired to construct a solenoid that has a resistance of 5.00 Ω and that produces a magnetic field at its center of 4.00 × 10–2 T when it carries a current of 4.00 A. The solenoid is to be constructed from copper wire having a diameter of 0.500 mm. If the radius of the solenoid is to be 1.00 cm, determine (a) the number of turns of wire needed and (b) the length the solenoid should have.
Answer:
Known:
Resistance, R = 5.00 Ω
carries current, i = 4.00 A
magnetic field, B = 4.00 × 10–2 T
From R = ρL/A , the required length of wire to be used is
L = RA/ρ = 5.00 Ω x [π x (0.500 x 10─3)2 m/4]/( 1.70 x 10─8 Ω.m) = 57.7 m
The total number of turns on the solenoid (i.e., the number of times this length of wire will go around a 1.00 cm radius cylinder) is
N = L/2πr = 57.7 m/(2π x 0.01 m) = 919 turn
(b) From B = μ0nI, the number of turns per unit length on the solenoid is
n = B/μ0I = 4.00 × 10–2 T/(4π x 10─7 T.m/A x 4.00 A) = 7960 turn/m
Thus, the required length of the solenoid is
L = N/n = 919 turn/(7960 turn/m) = 0.115 m = 115 mm
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