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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 :

Kazeemsodikisola

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

snehashish65

The required value of peak output voltage is 169.65 Volts.

Given data:

The frequency of ac generator is, f = 60 Hz.

The number of turns of coil is, n = 125 turns.

The area of each coil is, [tex]A =3.0 \times 10^{-2} \;\rm m^{2}[/tex].

The strength of magnetic field is, B = 0.12 T.

To start with this problem, we need to find the peak emf first. The peak output voltage is nothing but the value of this peak emf only. The expression for the  peak EMF is given as,

ε = —dΦ/dt

here,

Φ is magnetic flux and it is given as

Φ = NBA Cosθ

Here,

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

Where θ =ωt and ω = 2πf

Then, θ = 2πft

So, the expression for the 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

 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

Thus, we can conclude that the required value of peak output voltage is 169.65 Volts.

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