A 380 turn solenoid of length 32.0 cm and radius 3.10 cm carries a current of 4.80 A. Find the following.
(a) the magnetic field strength inside the coil at its midpoint
mT
(b) the magnetic flux through a circular cross-sectional area of the solenoid at its midpoint
T · m2
(4pi*10^-7)(4.8*380/.32)
=____mT
To find the magnetic field strength inside the coil at its midpoint, we can use the formula for the magnetic field inside a solenoid:
B = μ₀ * N * I / L
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 in the solenoid (380 turns),
I is the current flowing through the solenoid (4.80 A), and
L is the length of the solenoid (32.0 cm = 0.32 m).
Substituting the given values into the formula:
B = (4π × 10⁻⁷ T·m/A) * (380 turns) * (4.80 A) / (0.32 m)
Calculating the expression:
B ≈ 0.060 T (to three decimal places)
Therefore, the magnetic field strength inside the coil at its midpoint is approximately 0.060 T.
To find the magnetic flux through a circular cross-sectional area of the solenoid at its midpoint, we can use the formula:
Φ = B * A
Where:
Φ is the magnetic flux,
B is the magnetic field strength (0.060 T), and
A is the cross-sectional area of the solenoid.
The cross-sectional area of a circular solenoid is given by:
A = π * r²
Where:
r is the radius of the solenoid (3.10 cm = 0.031 m).
Substituting the given values into the formula:
A ≈ π * (0.031 m)²
Calculating the expression:
A ≈ 0.00302 m² (to five significant figures)
Finally, calculating the magnetic flux:
Φ ≈ (0.060 T) * (0.00302 m²)
Φ ≈ 1.81 × 10⁻³ T·m² (to three significant figures)
Therefore, the magnetic flux through a circular cross-sectional area of the solenoid at its midpoint is approximately 1.81 × 10⁻³ T·m².