Answer:
When the plane of the loop is perpendicular to the field lines, the magnetic flux through the loop is 6.284 x 10⁻³ wb.
Explanation:
The magnetic flux through the loop is the product of magnetic field strength, circular area of the loop and angle of inclination.
Ф = BAcosθ
When the plane of the loop is perpendicular to the field lines, θ = 0
Ф = BAcos0
Ф = BA
Where;
Ф is the magnetic flux through the loop
B is the magnetic field strength = 0.8 T
A is the circular area of the loop;
[tex]A =\frac{\pi D^2}{4} = \frac{\pi *0.1^2}{4} = 0.007855 m^2[/tex]
Ф = 0.8 x 0.007855 = 6.284 x 10⁻³ wb
Therefore, when the plane of the loop is perpendicular to the field lines, the magnetic flux through the loop is 6.284 x 10⁻³ wb.
0.006598wb
The magnetic flux,Ф, (measured in weber, wb) through a given loop of wire is related to the area (A) of the wire and the magnetic field (B) in which the loop is placed as follows;
Ф = BA cos θ -----------------------(i)
Where;
θ = angle between the planar area of the coil and the magnetic flux
A = π d² / 4 [d = diameter of the loop]
From the question;
B = 0.84T
θ = 0° [since the plane of the loop is perpendicular to the field lines, it makes it parallel to the magnetic flux]
d = 10cm = 0.1m
=> A = π (0.1)² / 4 [Take π = 3.142]
=> A = 3.142 x (0.1)² / 4
=> A = 0.007855m²
Substitute these values into equation (i) as follows;
Ф = 0.84 x 0.007855 x cos 0°
Ф = 0.84 x 0.007855 x 1
Ф = 0.006598
Therefore, the magnetic flux through the loop is 0.006598wb
