A Field Dying Away

solved FaradaysLaw

Q. A square loop of side 10 cm $A = 0.01$ m² and resistance 0.5 Ω $R = 0.5$ Ω is placed vertically in the east-west plane. A uniform magnetic field of 0.10 T $B_{0} = 0.10$ T is set up across the plane in the north-east direction $θ_{0} = 45 °$ , $θ_{1} = 45 °$ . The magnetic field is decreased to zero $B_{1} = 0$ T in 0.70 s $Δ t = 0.70$ s at a steady rate. What is the magnitude of the induced emf during this time-interval, in volts?

Inputs

resistance

R

0.5 Ω

field

B_{0}

0.1 T

time

Δ t

0.70 s

Outputs

emf

ε

How it's solved

Solved with FaradaysLaw

Given

$R = 0.5$ Ω
$B_{0} = 0.10$ T
$Δ t = 0.70$ s
$A = 0.01$ m² from “square loop of side 10 cm”
$N = 1$ from “A square loop”
$θ_{0} = 45 °$ from “in the north-east direction”
$θ_{1} = 45 °$ from “in the north-east direction”
$B_{1} = 0$ T from “decreased to zero”

Find $ε$ .

Working

flux through a loop(square, fading, 1) $Φ_{0} = B_{0} \cdot A \cdot \cos θ_{0}$
$Φ_{0} = 0.1 \cdot 0.01 \cdot \cos 45 °$
$Φ_{0} = 7.07106781 \times 10^{−4}$ Wb
flux through a loop(square, fading, 2) $Φ_{1} = B_{1} \cdot A \cdot \cos θ_{1}$
$Φ_{1} = 0 \cdot 0.01 \cdot \cos 45 °$
$Φ_{1} = 0$ Wb
emf from the change of flux(square, fading) $ε = \frac{N \cdot (Φ_{0} - Φ_{1})}{Δ t}$
$ε = \frac{1 \cdot (7.07106781 \times 10^{−4} - 0)}{0.7}$
$ε = 0.00101$ V
$ε \approx 0.0010$ Vto 2 significant figures

Conclusion

The field cuts the loop slantwise, so only its share along the loop's normal counts; as that share dies away evenly, the flux falls at a steady rate and drives a steady emf.

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_field_dying_away do4  kind :numerical5  describes "A square loop of side 10 cm and resistance %{resistance} is placed vertically in the east-west plane. A " \6            "uniform magnetic field of %{field} is set up across the plane in the north-east direction. The magnetic " \7            "field is decreased to zero in %{time} at a steady rate. %{emf} during this time-interval, in volts?"8  given resistance: "0.5 Ω"9  given field: "0.10 T"10  given time: "0.70 s"11  asks emf: "What is the magnitude of the induced emf"12end

solution.rb

1require_relative "../questions/a_field_dying_away"2require_relative "../science/electromagnetic_induction/faradays_law"34solve :a_field_dying_away, using: FaradaysLaw do5  objects :square6  square.goes(:fading)78  take resistance: square.resistance9  take field: square.magnetic_field(at: fading.start)10  take time: fading.time_taken1112  deduce_from "square loop of side 10 cm", square.area => "0.01 m2"13  deduce_from "A square loop", square.turns => "1"14  deduce_from "in the north-east direction", square.angle_to_the_field(at: fading.start) => "45 deg",15                                             square.angle_to_the_field(at: fading.end) => "45 deg"16  deduce_from "decreased to zero", square.magnetic_field(at: fading.end) => "0 T"1718  emf = square.induced_emf(during: fading)19  answer emf:, in: "V", figures: 22021  conclusion "The field cuts the loop slantwise, so only its share along the loop's normal counts; as that " \22             "share dies away evenly, the flux falls at a steady rate and drives a steady emf."23end

playground.rb

1require_relative "../solutions/a_field_dying_away"23play_with :a_field_dying_away do4  title "A Field Dying Away"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 square during fading	$Δ x$	m	—
position of square at end of fading	$x_{1}$	m	—
position of square at start of fading	$x_{0}$	m	—
average velocity of square during fading	$v_{av}$	m/s	—
time taken of fading	$Δ t$	s	0 … 1e+05
average speed of square during fading	$c_{av}$	m/s	—
path length of square during fading	$d$	m	—
magnetic flux of square at start of fading	$Φ_{0}$	Wb	—
magnetic field of square at start of fading	$B_{0}$	T	—
area of square	$A$	m²	—
angle to the field of square at start of fading	$θ_{0}$	—	0 … 3.142
magnetic flux of square at end of fading	$Φ_{1}$	Wb	—
magnetic field of square at end of fading	$B_{1}$	T	—
angle to the field of square at end of fading	$θ_{1}$	—	0 … 3.142
induced emf of square during fading	$ε$	V	—
turns of square	$N$	—	—
induced current of square during fading	$I$	A	—
resistance of square	$R$	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(fading)

v_{av} = \frac{Δ x}{Δ t}

displacement a time(fading)

c_{av} = \frac{d}{Δ t}

path a time(fading)

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

flux through a loop(square, fading, 1)

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

flux through a loop(square, fading, 2)

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

emf from the change of flux(square, fading)

I = \frac{ε}{R}

current driven by the emf(square, fading)