Question

A 10-cm-diameter circular loop of wire is placed in a 0.84-T magnetic field. Part A When the plane of the loop is perpendicular to the field lines, what is the magnetic flux through the loop?

Answer 1

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;

$A =\frac{\pi D^2}{4} = \frac{\pi *0.1^2}{4} = 0.007855 m^2$

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.

Answer 2

Answer:

0.006598wb

Explanation:

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