Question

The coil in a 60-Hz ac generator has 125 turns, each having an area of 3.0 × 10-2 m2 and is rotated in a uniform 0.12-T magnetic field. What is the peak output voltage of this generator?

Answer 1

Answer:

Explanation:

Given that, .

Frequency

f = 60Hz

Number of turns

N = 125turns

Surface area of coil

A = 3 × 10^-2 m²

Magnetic field

B = 0.12T

Voltage peak to peak? I.e the EMF

EMF is given as

ε = —dΦ/dt

Where Φ is magnetic flux and it is given as

Φ = NBA Cosθ

Where N is number of turns

B is magnetic field

A is the cross sectional area

And θ is the resulting angle from the dot product of area and magnetic field

Where θ =ωt and ω = 2πf

Then, θ = 2πft

So, your magnetic flux becomes

Φ = NBA Cos(2πft)

Now, dΦ / dt = —NBA•2πf Sin(2πft)

dΦ / dt = —2πf • NBA Sin(2πft)

So, ε = —dΦ/dt

Then,

ε = 2πf • NBA Sin(2πft)

So, the maximum peak to peak emf will occur when the sine function is 1

I.e Sin(2πft) = 1

So, the required peak to peak emf is

ε = 2πf • NBA

Substituting all the given parameters

ε = 2π × 60 × 125 × 0.12 × 3 × 10^-2

ε = 169.65 Volts

The peak to peak voltage is 169.65 V

Answer 2

Answer: 169.67 V

Explanation:

From Faraday's Law, we know that

emf = - d(magnetic flux) / d t

magnetic flux = N B A cos (θ), where

N = number of turns

B = magnetic field

A = area of coils

cos(θ) = results from dot product of magnetic field vector and area vector.

θ = a function of time as the coil is rotating.

emf = N B A (2πf) sin (2π f t)

f = frequency

sin 2πft = 1

The peak output voltage is then

emf = (125) * (0.12) * (3*10^-2) * (2 * 3.142 * 60)

emf = 0.45 * 377.04

emf = 169.67 V