A Field Dying Away

Q. A square loop of side 10 cm and resistance 0.5 Ω is placed vertically in the east-west plane. A uniform magnetic field of 0.10 T is set up across the plane in the north-east direction. The magnetic field is decreased to zero in 0.70 s at a steady rate. What is the magnitude of the induced emf during this time-interval, in volts? expects 0.0010 V

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The equations it used

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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?

The playground

Inputs

resistance

R

0.5 Ω

field

B_{0}

0.1 T

time

Δ t

0.70 s

Outputs

emf

ε

The solution

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 Ruby

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

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

1require_relative "../solutions/a_field_dying_away"23play_with :a_field_dying_away do4  title "A Field Dying Away"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 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)