Pedalling a Generator

solved ACGenerator

Q. Kamla peddles a stationary bicycle. The pedals of the bicycle are attached to a 100 $N = 100$ turn coil of area 0.10 m² $A = 0.10$ m². The coil rotates at 0.5 Hz $ν = 0.5$ Hz and it is placed in a uniform magnetic field of 0.01 T $B = 0.01$ T perpendicular to the axis of rotation of the coil. What is the maximum voltage generated in the coil, in volts?

Inputs

turns

N

100

area

A

0.1 m²

rate

ν

0.5 Hz

field

B

0.01 T

Outputs

peak

ε_{0}

How it's solved

Solved with ACGenerator

Given

$N = 100$
$A = 0.10$ m²
$ν = 0.5$ Hz
$B = 0.01$ T

Find $ε_{0}$ .

Working

peak emf of a turning coil(coil) $ε_{0} = N \cdot B \cdot A \cdot 2 \cdot π \cdot ν$
$ε_{0} = 100 \cdot 0.01 \cdot 0.1 \cdot 2 \cdot 3.14159265 \cdot 0.5$
$ε_{0} = 0.3142$ V
$ε_{0} \approx 0.314$ Vto 3 significant figures

Conclusion

The flux through the turning coil swings as a cosine, so the emf peaks at the turns times the field times the area times the angular speed.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :pedalling_a_generator do4  kind :numerical5  describes "Kamla peddles a stationary bicycle. The pedals of the bicycle are attached to a %{turns} turn coil of " \6            "area %{area}. The coil rotates at %{rate} and it is placed in a uniform magnetic field of %{field} " \7            "perpendicular to the axis of rotation of the coil. %{peak}, in volts?"8  given turns: "100"9  given area: "0.10 m2"10  given rate: "0.5 Hz"11  given field: "0.01 T"12  asks peak: "What is the maximum voltage generated in the coil"13end

solution.rb

1require_relative "../questions/pedalling_a_generator"2require_relative "../science/electromagnetic_induction/ac_generator"34solve :pedalling_a_generator, using: ACGenerator do5  objects :coil67  take turns: coil.turns8  take area: coil.area9  take rate: coil.frequency10  take field: B1112  peak = coil.peak_emf13  answer peak:, in: "V", figures: 31415  conclusion "The flux through the turning coil swings as a cosine, so the emf peaks at the turns times the " \16             "field times the area times the angular speed."17end

playground.rb

1require_relative "../solutions/pedalling_a_generator"23play_with :pedalling_a_generator do4  title "Pedalling a Generator"5end

The laws it drew on

quantity	written	units	within	from
pi	$π$	—	0 … ∞	ACGenerator
magnetic field	$B$	T	—	ACGenerator
turns	$N$	—	—	ACGenerator
area	$A$	m²	—	ACGenerator
frequency	$ν$	Hz	0 … ∞	ACGenerator
peak emf	$ε_{0}$	V	—	ACGenerator
peak emf of ground	$ε_{0,g}$	V	—
turns of ground	$N_{g}$	—	—
area of ground	$A_{g}$	m²	—
frequency of ground	$ν_{g}$	Hz	0 … ∞
peak emf of coil	$ε_{0}$	V	—
turns of coil	$N$	—	—
area of coil	$A$	m²	—
frequency of coil	$ν$	Hz	0 … ∞

ε_{0,c} = N_{c} \cdot B \cdot A_{c} \cdot 2 \cdot π \cdot ν_{c}

peak emf of a turning coil

ε_{0,g} = N_{g} \cdot B \cdot A_{g} \cdot 2 \cdot π \cdot ν_{g}

peak emf of a turning coil(ground)

ε_{0} = N \cdot B \cdot A \cdot 2 \cdot π \cdot ν

peak emf of a turning coil(coil)