A Cut Loop Drawn Out Endwise for How Long

Q. A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. The velocity of the loop is 1 cm/s in a direction normal to the shorter side of the loop. For how long does the induced voltage last, in seconds? expects 8.0 s

What the draft establishes

Fills in once a draft is checked.

The web of concepts

given read from the words equation asked

The equations it used

None yet.

Check an answer

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. For how long does the induced voltage last, in seconds?

The playground

Inputs

field

B_{0}

0.3 T

speed

c_{av}

1 cm/s

Outputs

time

Δ t

The solution

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 $Δ t$ .

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) $c_{av} = \frac{d}{Δ t}$
$0.01 = \frac{0.08}{Δ t}$
$Δ t = 8.00 s$
$Δ t \approx 8.0$ sto 2 significant figures

Conclusion

The emf lasts only while some of the loop is still in the field and the flux through it is changing, the time the loop takes to cross its own width.

The Ruby

1require_relative "../physics"23question :a_cut_loop_drawn_out_endwise_for_how_long 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. %{time}, in seconds?"8  given field: "0.3 T"9  given speed: "1 cm/s"10  asks time: "For how long does the induced voltage last"11end

1require_relative "../questions/a_cut_loop_drawn_out_endwise_for_how_long"2require_relative "../science/electromagnetic_induction/faradays_law"34solve :a_cut_loop_drawn_out_endwise_for_how_long, 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  time = leaving.time_taken19  answer time:, in: "s", figures: 22021  conclusion "The emf lasts only while some of the loop is still in the field and the flux through it is " \22             "changing, the time the loop takes to cross its own width."23end

1require_relative "../solutions/a_cut_loop_drawn_out_endwise_for_how_long"23play_with :a_cut_loop_drawn_out_endwise_for_how_long do4  title "A Cut Loop Drawn Out Endwise for How Long"5end

The laws

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)