Answer:
Φ = 2.06×10^-3weber
Explanation:
Magnetic flux density is the ratio of the flux links to the area of the coil.
This can be expressed as
B =Φ/NA where:
Φ is the magnetic flux
B is the magnetic field strength
A is the area of the coil.
N is the number of turn in the wire
Given
B = 2.95 mT = 2.95×10^-3T
N = 36
radius = 7.85cm = 0.0785m
A = πr² ( since it is a circular coil)
A = π(0.0785)²
A = 0.0194m²
From the formula Φ = BAN
Φ = 2.95×10^-3 × 0.0194×36
Φ = 2.06×10^-3weber
2.056 x 10⁻³ Wb
The magnetic flux, Φ₀, through a single turn of a coil is given by the dot product between the magnetic field, B, and the area, A, of the coil. i.e
Φ₀ = B.A cos θ --------------(i)
Where;
θ = angle between the area vector and the magnetic field.
But for a coil with N number of turns, the magnetic flux, Φ₀, is actually the product of the number of turns and the magnetic flux through its single turn, Φ₀. i.e
Φ = N Φ₀ --------------(ii)
Substitute the value of Φ₀ in equation (i) into equation (ii) as follows;
Φ = N BAcosθ --------------(iii)
From the question, the coil is perpendicular to the magnetic field. This implies that the area vector is parallel to the magnetic field. Therefore equation (iii) becomes;
Φ = N BA ------------------(iv)
Also,
A = πr²
Where;
r = radius of the coil
Substitute A = πr² into equation (iv) as follows;
Φ = N B(πr²)
Φ = Nπr²B --------------(v)
From the question;
N = 36.0
r = 7.85cm = 0.0785m
B = 2.95mT = 2.95 x 10⁻³ T
Substitute these values into equation (v) as follows and take π = 3.142;
Φ = 36.0 x 3.142 x 0.0785² x 2.95 x 10⁻³
Φ = 0.002056
Φ = 2.056 x 10⁻³ Wb
Therefore, the magnetic flux linking its turns is 2.056 x 10⁻³ Wb
