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3 years ago

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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?

2 answers:

Ivanshal [37]3 years ago

7 0

The frequency of the generator is about 7.1 Hz

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<h3>Further explanation</h3>

<em>Let's recall </em><em>eletromotive force</em><em> formula for a generator:</em>

$\large {\boxed {\varepsilon = NBA\omega \sin (\omega t) }$

<em>ε = electromotive force ( V )</em>

<em>B = magnetic field strength (T)</em>

<em>N= number of turns in a coil</em>

<em>A = cross-sectional area ( m² )</em>

<em>ω = angular frequency ( rad/s )</em>

<em>t = time taken ( s )</em>

Let's tackle the problem now !

$\texttt{ }$

<u>Given:</u>

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

<u>Asked:</u>

frequency of the generator = f = ?

<u>Solution:</u>

$\varepsilon = NBA\omega \sin (\omega t)$

$\varepsilon_{max} = NBA\omega$

$\varepsilon_{max} = NBA(2 \pi f)$

$f = \varepsilon_{max} \div ( 2\pi NBA )$

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

$f = 8.0 \div [ 0.36 \pi ]$

$f = \frac{200}{9 \pi}$

$f \approx 7.1 \texttt{ Hz}$

$\texttt{ }$

<h3>Learn more</h3>

Temporary and Permanent Magnet : brainly.com/question/9966993
The three resistors : brainly.com/question/9503202
A series circuit : brainly.com/question/1518810
Compare and contrast a series and parallel circuit : brainly.com/question/539204

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<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Magnetic Field

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

shutvik [7]3 years ago

4 0

Answer:

7.1 Hz

Explanation:

In a generator, the maximum induced emf is given by

$\epsilon= 2\pi NAB f$

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

$A=0.030 m^2$

$\epsilon=8.0 V$

B = 0.030 T

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

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

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