The coil of a generator has a radius of 0.12 m. When this coil is unwound, the wire from which it is made has a length of 5.8 m. The magnetic field of the generator is 0.21 T, and the coil rotates at an angular speed of 26 rad/s. What is the peak emf of this generator?

Answer in V

To find the peak emf (electromotive force) of the generator, we can use Faraday's law of electromagnetic induction. According to this law, the emf induced in a coil is equal to the rate of change of magnetic flux through the coil.

The magnetic flux through a coil is given by the product of the magnetic field strength (B), the area of the coil (A), and the cosine of the angle between the magnetic field and the normal to the coil (θ). In this case, the angle is 90 degrees because the magnetic field is perpendicular to the plane of the coil.

So, the magnetic flux (Φ) can be calculated as:

Φ = B * A * cos(θ)

Now, let's find the area of the coil. The coil is unwound, so its shape is like a rectangle. The length of the wire from which the coil is made determines the length of the rectangle, and the radius of the coil determines the width of the rectangle.

The length of the rectangle (l) is equal to the length of the wire, which is given as 5.8 m.

The width of the rectangle (w) is equal to the circumference of the circle, which can be calculated as 2π * r, where r is the radius of the coil given as 0.12 m.

So, the area (A) of the coil is given by:

A = l * w
A = 5.8 m * 2π * 0.12 m
A ≈ 1.1 m²

Now, let's substitute the given values into the formula for magnetic flux:

Φ = B * A * cos(θ)
Φ = 0.21 T * 1.1 m² * cos(90 degrees)
Φ ≈ 0.21 T * 1.1 m² * 0
Φ ≈ 0

Since the magnetic flux is zero, there is no change in flux with time. Hence, there is no induced emf.

Therefore, the peak emf of this generator is zero volts (0 V).