A Cut Loop Drawn Out Endwise

solved FaradaysLaw

Q. A rectangular wire loop $N = 1$ of sides 8 cm and 2 cm $A = 16$ cm² with a small cut is moving out of a region $B_{1} = 0$ T of uniform magnetic field of magnitude 0.3 T $B_{0} = 0.3$ T directed normal to the loop $θ_{0} = 0 °$ , $θ_{1} = 0 °$ . The velocity of the loop is 1 cm/s $c_{av} = 1$ cm/s in a direction normal to the shorter side $d = 8$ cm of the loop. What is the emf developed across the cut, in volts?

Inputs

field

B_{0}

0.3 T

speed

c_{av}

1 cm/s

Outputs

emf

ε

How it's solved

Solved with FaradaysLaw

Given

$B_{0} = 0.3$ T
$c_{av} = 1$ cm/s
$A = 16$ cm² from “sides 8 cm and 2 cm”
$N = 1$ from “A rectangular wire loop”
$θ_{0} = 0 °$ from “directed normal to the loop”
$θ_{1} = 0 °$ from “directed normal to the loop”
$B_{1} = 0$ T from “moving out of a region”
$d = 8$ cm from “normal to the shorter side”

Find $ε$ .

Working

$c_{av} = 1$ cm/s $= 0.01$ m/s
$A = 16$ cm² $= 0.0016$ m²
$d = 8$ cm $= 0.08$ m
path a time(leaving) $c_{av} = \frac{d}{Δ t}$
$0.01 = \frac{0.08}{Δ t}$
$Δ t = 8$ s
flux through a loop(rectangle, leaving, 1) $Φ_{0} = B_{0} \cdot A \cdot \cos θ_{0}$
$Φ_{0} = 0.3 \cdot 0.0016 \cdot \cos 0 °$
$Φ_{0} = 4.8 \times 10^{−4}$ Wb
flux through a loop(rectangle, leaving, 2) $Φ_{1} = B_{1} \cdot A \cdot \cos θ_{1}$
$Φ_{1} = 0 \cdot 0.0016 \cdot \cos 0 °$
$Φ_{1} = 0$ Wb
emf from the change of flux(rectangle, leaving) $ε = \frac{N \cdot (Φ_{0} - Φ_{1})}{Δ t}$
$ε = \frac{1 \cdot (4.8 \times 10^{−4} - 0)}{8}$
$ε = 6 \times 10^{−5}$ V
$ε \approx 6.0 \times 10^{−5}$ Vto 2 significant figures

Conclusion

The flux through the loop falls steadily as it slides out, so the emf across the cut is the whole flux shared over the time the loop takes to leave.

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_cut_loop_drawn_out_endwise do4  kind :numerical5  describes "A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform " \6            "magnetic field of magnitude %{field} directed normal to the loop. The velocity of the loop is " \7            "%{speed} in a direction normal to the shorter side of the loop. %{emf}, in volts?"8  given field: "0.3 T"9  given speed: "1 cm/s"10  asks emf: "What is the emf developed across the cut"11end

solution.rb

1require_relative "../questions/a_cut_loop_drawn_out_endwise"2require_relative "../science/electromagnetic_induction/faradays_law"34solve :a_cut_loop_drawn_out_endwise, using: FaradaysLaw do5  objects :rectangle6  rectangle.goes(:leaving)78  take field: rectangle.magnetic_field(at: leaving.start)9  take speed: rectangle.average_speed(during: leaving)1011  deduce_from "sides 8 cm and 2 cm", rectangle.area => "16 cm2"12  deduce_from "A rectangular wire loop", rectangle.turns => "1"13  deduce_from "directed normal to the loop", rectangle.angle_to_the_field(at: leaving.start) => "0 deg",14                                             rectangle.angle_to_the_field(at: leaving.end) => "0 deg"15  deduce_from "moving out of a region", rectangle.magnetic_field(at: leaving.end) => "0 T"16  deduce_from "normal to the shorter side", rectangle.path_length(during: leaving) => "8 cm"1718  emf = rectangle.induced_emf(during: leaving)19  answer emf:, in: "V", figures: 22021  conclusion "The flux through the loop falls steadily as it slides out, so the emf across the cut is the " \22             "whole flux shared over the time the loop takes to leave."23end

playground.rb

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

v_{av} = \frac{Δ x}{Δ t}

displacement a time(leaving)

c_{av} = \frac{d}{Δ t}

path a time(leaving)

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

flux through a loop(rectangle, leaving, 1)

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

flux through a loop(rectangle, leaving, 2)

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

emf from the change of flux(rectangle, leaving)

I = \frac{ε}{R}

current driven by the emf(rectangle, leaving)