A Coil Turned Over

solved FaradaysLaw

Q. A circular coil of radius 10 cm $A = 0.0314159$ m², 500 $N = 500$ turns and resistance 2 Ω $R = 2$ Ω is placed with its plane perpendicular to the horizontal component $θ_{0} = 0 °$ of the earth's magnetic field. It is rotated about its vertical diameter through 180° $θ_{1} = 180 °$ , $B_{1} = B_{0}$ in 0.25 s $Δ t = 0.25$ s. The horizontal component of the earth's magnetic field at the place is 3.0 × 10⁻⁵ T $B_{0} = 3.0 \times 10^{−5}$ T. What is the magnitude of the emf induced in the coil, in volts?

Inputs

turns

N

500

resistance

R

2 Ω

time

Δ t

0.25 s

field

B_{0}

3e-05 T

Outputs

emf

ε

How it's solved

Solved with FaradaysLaw

Given

$N = 500$
$R = 2$ Ω
$Δ t = 0.25$ s
$B_{0} = 3.0 \times 10^{−5}$ T
$A = 0.0314159$ m² from “A circular coil of radius 10 cm”
$θ_{0} = 0 °$ from “plane perpendicular to the horizontal component”
$θ_{1} = 180 °$ from “through 180°”
$B_{1} = B_{0}$ from “through 180°”

Find $ε$ .

Working

flux through a loop(coil, turning, 1) $Φ_{0} = B_{0} \cdot A \cdot \cos θ_{0}$
$Φ_{0} = 3 \times 10^{−5} \cdot 0.0314159 \cdot \cos 0 °$
$Φ_{0} = 9.42477 \times 10^{−7}$ Wb
through 180° $B_{1} = B_{0}$
$B_{1} = 3 \times 10^{−5}$
$B_{1} = 3 \times 10^{−5}$ T
flux through a loop(coil, turning, 2) $Φ_{1} = B_{1} \cdot A \cdot \cos θ_{1}$
$Φ_{1} = 3 \times 10^{−5} \cdot 0.0314159 \cdot \cos 180 °$
$Φ_{1} = −9.42477 \times 10^{−7}$ Wb
emf from the change of flux(coil, turning) $ε = \frac{N \cdot (Φ_{0} - Φ_{1})}{Δ t}$
$ε = \frac{500 \cdot (9.42477 \times 10^{−7} - (−9.42477 \times 10^{−7}))}{0.25}$
$ε = 0.00377$ V
$ε \approx 0.0038$ Vto 2 significant figures

Conclusion

Turning the coil over reverses the flux through it, so the change is twice the flux, shared over every turn of the coil.

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 :a_coil_turned_over do4  kind :numerical5  describes "A circular coil of radius 10 cm, %{turns} turns and resistance %{resistance} is placed with its plane " \6            "perpendicular to the horizontal component of the earth's magnetic field. It is rotated about its " \7            "vertical diameter through 180° in %{time}. The horizontal component of the earth's magnetic field at " \8            "the place is %{field}. %{emf}, in volts?"9  given turns: "500"10  given resistance: "2 Ω"11  given time: "0.25 s"12  given field: "3.0e-5 T"13  asks emf: "What is the magnitude of the emf induced in the coil"14end

solution.rb

1require_relative "../questions/a_coil_turned_over"2require_relative "../science/electromagnetic_induction/faradays_law"34solve :a_coil_turned_over, using: FaradaysLaw do5  objects :coil6  coil.goes(:turning)78  take turns: coil.turns9  take resistance: coil.resistance10  take time: turning.time_taken11  take field: coil.magnetic_field(at: turning.start)1213  deduce_from "A circular coil of radius 10 cm", coil.area => "0.0314159 m2"14  deduce_from "plane perpendicular to the horizontal component", coil.angle_to_the_field(at: turning.start) => "0 deg"15  deduce_from "through 180°", coil.angle_to_the_field(at: turning.end) => "180 deg",16                              coil.magnetic_field(at: turning.end) => coil.magnetic_field(at: turning.start)1718  emf = coil.induced_emf(during: turning)19  answer emf:, in: "V", figures: 22021  conclusion "Turning the coil over reverses the flux through it, so the change is twice the flux, shared " \22             "over every turn of the coil."23end

playground.rb

1require_relative "../solutions/a_coil_turned_over"23play_with :a_coil_turned_over do4  title "A Coil Turned Over"5end

The laws it drew on

quantity	written	units	within	from
position	$x$	m	—	Velocity
displacement	$Δ x$	m	—	Velocity
path length	$d$	m	—	Velocity
time taken	$Δ t$	s	0 … 1e+05	Velocity
average velocity	$v_{av}$	m/s	—	Velocity
average speed	$c_{av}$	m/s	—	Velocity
starting position	$x_{0}$	m	—	Velocity
final position	$x$	m	—	Velocity
turns	$N$	—	—	FaradaysLaw
area	$A$	m²	—	FaradaysLaw
resistance	$R$	Ω	0 … ∞	FaradaysLaw
magnetic field	$B$	T	—	FaradaysLaw
angle to the field	$θ$	—	0 … 3.142	FaradaysLaw
magnetic flux	$Φ$	Wb	—	FaradaysLaw
induced emf	$ε$	V	—	FaradaysLaw
induced current	$I$	A	—	FaradaysLaw
displacement of coil during turning	$Δ x$	m	—
position of coil at end of turning	$x_{1}$	m	—
position of coil at start of turning	$x_{0}$	m	—
average velocity of coil during turning	$v_{av}$	m/s	—
time taken of turning	$Δ t$	s	0 … 1e+05
average speed of coil during turning	$c_{av}$	m/s	—
path length of coil during turning	$d$	m	—
magnetic flux of coil at start of turning	$Φ_{0}$	Wb	—
magnetic field of coil at start of turning	$B_{0}$	T	—
area of coil	$A$	m²	—
angle to the field of coil at start of turning	$θ_{0}$	—	0 … 3.142
magnetic flux of coil at end of turning	$Φ_{1}$	Wb	—
magnetic field of coil at end of turning	$B_{1}$	T	—
angle to the field of coil at end of turning	$θ_{1}$	—	0 … 3.142
induced emf of coil during turning	$ε$	V	—
turns of coil	$N$	—	—
induced current of coil during turning	$I$	A	—
resistance of coil	$R$	Ω	0 … ∞

x_{0} \equiv x_{i} (start of motion)

starting position

x \equiv x_{i} (end of motion)

final position

d_{i} (whole) = d_{i} (part 1) + d_{i} (part 2)

path in parts

Δ t_{w} = Δ t_{p} + Δ t_{p}

time in parts

Δ x = x_{1} - x_{0}

change of position(turning)

v_{av} = \frac{Δ x}{Δ t}

displacement a time(turning)

c_{av} = \frac{d}{Δ t}

path a time(turning)

Φ_{0} = B_{0} \cdot A \cdot \cos θ_{0}

flux through a loop(coil, turning, 1)

Φ_{1} = B_{1} \cdot A \cdot \cos θ_{1}

flux through a loop(coil, turning, 2)

ε = \frac{N \cdot (Φ_{0} - Φ_{1})}{Δ t}

emf from the change of flux(coil, turning)

I = \frac{ε}{R}

current driven by the emf(coil, turning)

B_{1} = B_{0}

through 180°