A 14 cm diameter circular loop of wire is placed in a 0.58 T magnetic field.

(a) When the plane of the loop is perpendicular to the field lines, what is the magnetic flux through the loop?
(b) The plane of the loop is rotated until it makes a 36° angle with the field lines. What is the angle è in Eq. 21-1 for this situation?
(c) What is the magnetic flux through the loop at this angle?

(a) Flux = (loop area) x (magnetic field. The area must be in square meters. The field must be in Tesla. The units of the flux are called Webers.

(b) You have not defined the angle "è". If it is the angle between the B field lines and the normal to the loop, that angle is 54 degrees.
(c) Take your answer to (a) and multiply it by sin 36.

To answer these questions, we need to use the formula for magnetic flux:

Φ = B * A * cos(θ)

where:
- Φ is the magnetic flux
- B is the magnetic field strength
- A is the area of the loop
- θ is the angle between the magnetic field lines and the normal to the loop

Let's break down each part of the question.

(a) When the plane of the loop is perpendicular to the field lines, the angle θ is 0°. We are given the diameter of the loop, which is 14 cm. To find the area of the loop, we need to calculate the radius and then use the formula for the area of a circle:

A = π * r^2

where r is the radius of the loop. The radius is half the diameter, so r = 14 cm / 2 = 7 cm = 0.07 m.

Plugging in the values, we have:
Φ = B * A * cos(0°)
= 0.58 T * π * (0.07 m)^2 * cos(0) (cos(0) = 1)
= 0.036 T*m^2

Therefore, when the plane of the loop is perpendicular to the field lines, the magnetic flux through the loop is 0.036 T*m^2.

(b) When the plane of the loop is rotated until it makes a 36° angle with the field lines, we are asked to find the angle θ in the formula Φ = B * A * cos(θ). The given angle è corresponds to the angle between the field lines and the normal to the loop, but we need the angle θ, which is the angle between the field lines and the plane of the loop.

To find θ, we can subtract the given angle è from 90° (because the normal to the loop is perpendicular to the plane of the loop). Therefore, θ = 90° - 36° = 54°.

(c) Using the same formula for magnetic flux and the angle θ we just calculated, we can find the magnetic flux through the loop at this angle:

Φ = B * A * cos(θ)
= 0.58 T * π * (0.07 m)^2 * cos(54°)

Now you can plug these values into a calculator to find the magnetic flux through the loop at this angle.

To solve these questions, we will use the formula for magnetic flux:

ϕ = B * A * cos(θ)

Where:
- ϕ is the magnetic flux
- B is the magnetic field strength
- A is the area of the loop
- θ is the angle between the magnetic field lines and the normal to the loop

Let's solve these step by step:

(a) When the plane of the loop is perpendicular to the field lines, the angle θ is 90 degrees. The formula for magnetic flux simplifies to:

ϕ = B * A * cos(90°)

Since cos(90°) = 0, the magnetic flux through the loop is 0.

(b) When the plane of the loop is rotated until it makes a 36° angle with the field lines, the angle θ is 36 degrees. To find the value of θ in Eq. 21-1, which is the rotational angle of the loop from the y-axis, we subtract 36 degrees from 90 degrees:

è = 90° - 36°
è = 54°

So, the angle è in Eq. 21-1 is 54°.

(c) To find the magnetic flux through the loop at this angle, we substitute the given values into the formula for magnetic flux:

ϕ = B * A * cos(θ)

Where:
- B = 0.58 T (given)
- A = π * r^2 (the area of a circle with radius r)

Since the diameter of the loop is 14 cm, the radius is half of that, which is 7 cm (or 0.07 m).

ϕ = 0.58 T * π * (0.07 m)^2 * cos(36°)

Using a calculator:

ϕ ≈ 0.0585 Wb

So, the magnetic flux through the loop at this angle is approximately 0.0585 Weber.