Problem#1
A solenoid that is 95.0 cm long has a radius of 2.00 cm and a winding of 1200 turns; it carries a current of 3.60 A. Calculate the magnitude of the magnetic field inside the solenoid.Answer:
Known:
Length solenoid, L = 95.0 cm
radius, r = 2.00 cm
number of turns, N = 1200
current, i = 3.60 A
B does not depend on radius r of the solenoid. So,
B = μ0ni = μ0Ni/L
B = (4π x 10─7 T.m/A)(3.60 A)(1200)/(0.95 m) = 5.71 x 10─3 T
Problem#2
A 200-turn solenoid having a length of 25 cm and a diameter of 10 cm carries a current of 0.29 A. Calculate the magnitude of the magnetic field B inside the solenoid.
Answer:
Known:
length solenoid, L = 25 cm
diameter, d = 10 cm
current, i = 0.29 A
number of turns, N = 200
B does not depend on radius r of the solenoid. So,
B = μ0ni = μ0Ni/L
B = (4π x 10─7 T.m/A)(0.29 A)(200)/(0.25 m) = 2.92 x 10─4 T
Problem#3
A solenoid 1.30 m long and 2.60 cm in diameter carries a current of 18.0 A. The magnetic field inside the solenoid is 23.0 mT. Find the length of the wire forming the solenoid.
Answer:
Known:
length solenoid, L = 1.30 m
diameter, d = 2.60 cm
current, i = 18.0 A
magnetic field inside the solenoid, B = 23.0 mT = 0.023 T
We use equations B = μ0ni = μ0Ni/L to determine the number of turns. Therefore,
0.023 T = (4π x 10─7 T.m/A)(18.0 A)N/(1.30 m)
N ≈ 1,322 loops
So, length of wire is
Lw = πdN = π x 0.026 m x 1,322 = 107.98 m
Problem#4
A long solenoid has 100 turns/cm and carries current i. An electron moves within the solenoid in a circle of radius 2.30 cm perpendicular to the solenoid axis. The speed of the electron is 0.0460c (c = speed of light). Find the current i in the solenoid.
Answer:
Known:
number of turns, n = 100/cm = 10000/m
radius, r = 2.30 cm
speed of the electron, v = 0.0460c
the magnetic field inside the solenoid is parallel to the axis of the solenoid, and the motion of the electron is perpendicular to the magnetic field which means the electron moves in a circular motion with a radius of r, this radius is given by:
r = mv/eB or B = mv/er
subtitute with the givens to get:
B = (9.11 x 10─31 kg)(0.0460 x 8 x 108 m/s)/(1.6 x 10─19 C x 0.023 m)
B = 3.41 x 10─3 T
The magnetic field due to a solenoid is given by:
B = μ0nI
Where n is the number of turn per meter. The current is
I = B/μ0n = 3.41 x 10─3 T/(10000/m x 4π x 10─7 T.m/A) = 0.27 A
Problem#4
A long solenoid has 100 turns/cm and carries current i. An electron moves within the solenoid in a circle of radius 2.30 cm perpendicular to the solenoid axis. The speed of the electron is 0.0460c (c = speed of light). Find the current i in the solenoid.
Answer:
Known:
number of turns, n = 100/cm = 10000/m
radius, r = 2.30 cm
speed of the electron, v = 0.0460c
the magnetic field inside the solenoid is parallel to the axis of the solenoid, and the motion of the electron is perpendicular to the magnetic field which means the electron moves in a circular motion with a radius of r, this radius is given by:
r = mv/eB or B = mv/er
subtitute with the givens to get:
B = (9.11 x 10─31 kg)(0.0460 x 8 x 108 m/s)/(1.6 x 10─19 C x 0.023 m)
B = 3.41 x 10─3 T
The magnetic field due to a solenoid is given by:
B = μ0nI
Where n is the number of turn per meter. The current is
I = B/μ0n = 3.41 x 10─3 T/(10000/m x 4π x 10─7 T.m/A) = 0.27 A
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