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You are designing a generator with a maximum emf 8.0 V. If the generator coil has 200 turns and a cross-sectional area of 0.030 m2, what would be the frequency of the generator in a magnetic field of 0.030 T?

Respuesta :

skyluke89

Answer:

7.1 Hz

Explanation:

In a generator, the maximum induced emf is given by

[tex]\epsilon= 2\pi NAB f[/tex]

where

N is the number of turns in the coil

A is the area of the coil

B is the magnetic field strength

f is the frequency

In this problem, we have

N = 200

[tex]A=0.030 m^2[/tex]

[tex]\epsilon=8.0 V[/tex]

B = 0.030 T

So we can re-arrange the equation to find the frequency of the generator:

[tex]f=\frac{\epsilon}{2\pi NAB}=\frac{8.0 V}{2\pi (200)(0.030 m^2)(0.030 T)}=7.1 Hz[/tex]

johanrusli

The frequency of the generator is about 7.1 Hz

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Further explanation

Let's recall eletromotive force formula for a generator:

[tex]\large {\boxed {\varepsilon = NBA\omega \sin (\omega t) }[/tex]

ε = electromotive force ( V )

B = magnetic field strength (T)

N= number of turns in a coil

A = cross-sectional area ( m² )

ω = angular frequency ( rad/s )

t = time taken ( s )

Let's tackle the problem now !

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Given:

maximum emf = ε = 8.0 V

number of turns = N = 200 turns

cross-sectional area = A = 0.030 m²

magnetic field strength = B = 0.030 T

Asked:

frequency of the generator = f = ?

Solution:

[tex]\varepsilon = NBA\omega \sin (\omega t)[/tex]

[tex]\varepsilon_{max} = NBA\omega[/tex]

[tex]\varepsilon_{max} = NBA(2 \pi f)[/tex]

[tex]f = \varepsilon_{max} \div ( 2\pi NBA )[/tex]

[tex]f = 8.0 \div [ 2\pi (200)(0.030)(0.030) ][/tex]

[tex]f = 8.0 \div [ 0.36 \pi ][/tex]

[tex]f = \frac{200}{9 \pi}[/tex]

[tex]f \approx 7.1 \texttt{ Hz}[/tex]

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Answer details

Grade: High School

Subject: Physics

Chapter: Magnetic Field

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Keywords: Magnet , Field , Magnetic , Current , Wire , Unit ,

Ver imagen johanrusli