A cirular coil of 19 turns and radius 10.2cm surrounds a long solenoid of radius 2.6cm and 1.00x103 turns per meter. The two are oriented so that the axis of the coil and the axis of the solenoid are parallel and the center of the coil is at the center of the solenoid. If the current in the solenoid is 3.6A, What is the flux through the 19-turn coil? Answer in units of Webers.
To find the flux through the 19-turn coil, we need to apply Ampere's Law.
Ampere's Law states that the magnetic field (B) around a closed loop is equal to the product of the permeability of free space (μ₀) and the current enclosed (I), divided by the length of the path (l) integrated over the loop:
∮ B·dl = μ₀I
The flux through the coil can be calculated by integrating the magnetic field produced by the solenoid over the surface of the coil. Since the coil and solenoid are parallel, this simplifies the calculation.
First, let's find the magnetic field produced by the solenoid at the center of the coil.
The magnetic field inside a solenoid is given by the formula:
B = μ₀ * N * I
where:
B is the magnetic field strength
μ₀ is the permeability of free space (4π × 10⁻⁷ T·m/A)
N is the number of turns per meter
I is the current in the solenoid
Plugging in the given values:
N = 1.00 × 10³ turns/m
I = 3.6 A
B = (4π × 10⁻⁷ T·m/A) * (1.00 × 10³ turns/m) * (3.6 A)
B = 4.56 × 10⁻³ T
Since the coil is at the center of the solenoid, the magnetic field is the same at the center of the coil. Thus, the magnetic field inside the coil is 4.56 × 10⁻³ T.
Now, let's calculate the flux through the 19-turn coil. The flux (Φ) is defined as the product of the magnetic field (B) and the area (A) perpendicular to the magnetic field:
Φ = B * A
The area of a circular coil is given by the formula:
A = π * r²
where:
A is the area
π is a mathematical constant (approximately 3.14159)
r is the radius of the coil
Plugging in the given values:
r = 10.2 cm = 0.102 m
A = π * (0.102 m)²
A = 0.0328 m²
Φ = (4.56 × 10⁻³ T) * (0.0328 m²)
Φ = 1.50 × 10⁻⁴ Wb (Webers)
Therefore, the flux through the 19-turn coil is 1.50 × 10⁻⁴ Webers.