Find the magnetic flux through a solenoid of length 25cm, radius 1cm and 400turns , that carries a current of 3A
To find the magnetic flux through a solenoid, we can use the formula:
Φ = μ₀ * n * I * A
Where:
Φ is the magnetic flux
μ₀ is the permeability of free space (4π × 10^-7 T m/A)
n is the number of turns per unit length (turns/m)
I is the current flowing through the solenoid (A)
A is the cross-sectional area of the solenoid (m²)
Given:
Length (L) of the solenoid = 25 cm = 0.25 m
Radius (r) of the solenoid = 1 cm = 0.01 m
Number of turns (N) = 400 turns
Current (I) = 3 A
First, let's calculate the cross-sectional area A of the solenoid:
A = π * r^2
= π * (0.01 m)^2
= 0.000314 m²
Next, we need to find the number of turns per unit length n:
n = N / L
= 400 turns / 0.25 m
= 1600 turns/m
Finally, we can calculate the magnetic flux Φ:
Φ = μ₀ * n * I * A
= (4π × 10^-7 T m/A) * (1600 turns/m) * (3 A) * (0.000314 m²)
≈ 0.603 × 10^-3 T m²
Therefore, the magnetic flux through the solenoid is approximately 0.603 × 10^-3 T m².