A solenoid with a turn density of 300 turns/m, and radius, r = 18.0 cm, is oriented perpendicular to a 250-turn square coil with side-length, a = 15 cm as shown in the figure. The current through the solenoid varies in time as shown in the graph, with CCW as positive. №=250 I (A) SA/1.255 Top View 2. 6.0 4.0 2.0 0.0 -2.0 -4.0 -6.0 (a) (b) сси out 1.0 CCW I (.75 s) Solenoid Current vs. Time 2.0 = - -94 1.5 sec CW (in) 3.0 4.0 $₁ = B₂RH² = M₂ MIRR², B Ein = 3 Assec time (s) -Ɑ Calculate the magnetic flux through the square coil at t = 0.75 s and t = 3.1 s and indicate which way it flows through the square coil. N = = - Mon NT² I dt 5.0 C a 3A, B₁ = M₁ NI = I (3.)) = 4A, B₂ = Mo nI = (4x x 109 T. M/A) (300 /m ) ( TA) = (.5 ml (in) $(.15 5) = B₁ 24² = (1.13 x 10³ 7) * (₁18 m) = . 115 mub (out) (3.15) = 13₂x² = (1.5 X 102 t)^(.18 m)² = (53 Awb (a) I Solenoid Use Faraday's law to obtain a symbolic expression for the induced emf through the coil. - Montp² 1/11 (4π x 109 T-M/H) (300 /m) (3 A) = 1.13 mT (out) Calculate the slope of the current through the solenoid at t = 0.23 s, 1.84 s, and 4.3 s. For each time, indicate the direction of the current and the magnetic field it produces.

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2.
A solenoid with a turn density of 300 turns/m, and radius, r = 18.0 cm, is oriented
perpendicular to a 250-turn square coil with side-length, a = 15 cm as shown in the figure.
The current through the solenoid varies in time as shown in the graph, with CCW as positive.
N=250
I (A)
SA/1.255
Top View
Solenoid Current vs. Time
6.0
4.0
2.0
0.0
-2.0
-4.0
-6.0
(a)
CCW
out
1.0
сси
2.0
-9A.
33
T.Esec
3.0
4.0
3
b5sec
5.0
a
N +
=
| 112 = - Mon N T W ² 1 1 2
I
a
time (s)
Calculate the magnetic flux through the square coil at t = 0.75 s and t = 3.1 s and indicate
which way it flows through the square coil.
I (.75 s)
I (3.1)
:
$(155)=B₁ ap² = (1.13 × 10³ 7 ) x (₁ 184) = 115 mub (out)
§ (3.15) = 13₂ xf² = (1.5 x 102 t)^(-18 m)² = -(53 Awb (ñ)
r
Solenoid
3A, B₁ = M₁ 12 =
0
4A , B₂² Mo nI = (4x × 109 T.m/A) (300/m) (FA) =(.5 ml (in)
(4π *109 T-M/H) (300 /m) (3 A) = 1.13 mT (out)
(b)
Use Faraday's law to obtain a symbolic expression for the induced emf through the coil.
$₁ = B₂X² = M₂ MINH², doo - Mount the
13
Jt
Ein =
Calculate the slope of the current through the solenoid at t = 0.23 s, 1.84 s, and 4.3 s. For
each time, indicate the direction of the current and the magnetic field it produces.
Transcribed Image Text:2. A solenoid with a turn density of 300 turns/m, and radius, r = 18.0 cm, is oriented perpendicular to a 250-turn square coil with side-length, a = 15 cm as shown in the figure. The current through the solenoid varies in time as shown in the graph, with CCW as positive. N=250 I (A) SA/1.255 Top View Solenoid Current vs. Time 6.0 4.0 2.0 0.0 -2.0 -4.0 -6.0 (a) CCW out 1.0 сси 2.0 -9A. 33 T.Esec 3.0 4.0 3 b5sec 5.0 a N + = | 112 = - Mon N T W ² 1 1 2 I a time (s) Calculate the magnetic flux through the square coil at t = 0.75 s and t = 3.1 s and indicate which way it flows through the square coil. I (.75 s) I (3.1) : $(155)=B₁ ap² = (1.13 × 10³ 7 ) x (₁ 184) = 115 mub (out) § (3.15) = 13₂ xf² = (1.5 x 102 t)^(-18 m)² = -(53 Awb (ñ) r Solenoid 3A, B₁ = M₁ 12 = 0 4A , B₂² Mo nI = (4x × 109 T.m/A) (300/m) (FA) =(.5 ml (in) (4π *109 T-M/H) (300 /m) (3 A) = 1.13 mT (out) (b) Use Faraday's law to obtain a symbolic expression for the induced emf through the coil. $₁ = B₂X² = M₂ MINH², doo - Mount the 13 Jt Ein = Calculate the slope of the current through the solenoid at t = 0.23 s, 1.84 s, and 4.3 s. For each time, indicate the direction of the current and the magnetic field it produces.
(e)
Mo= 4x X107 T.M/H
Calculate the induced emf through the square coil at t = 0.8 s, 1.8 s, 2.8 s, and 3.8 s.
(Eand) - Montr²N #1
= (4x × 10^9 T-M/A) (30%/m) x (0.18 m)" (250) do
X
Calculate the induced current
through the square coil, which
has a total resistance of 1.28 ,
as needed to sketch a graph of
induced current vs. time, with
numerical labels.
Take CCW current as positive.
8-1.255= current ↑
flux point of
Induced flux
Induced current e cu
1.25s 2.955: Current
change fen to
at constant rate
Iind (mA)
L
1.0
(
(
flux change from
:
0 to Ⓡ
induced thy is
induced current e cw
Slope?
GP #4 EM Induction
Induced Current vs. Time
2.0
3.0
4.0
→
5.0
time (s)
I changes from G
flex change from x to.
induced Phux is → induced current
5 of 6
си
Transcribed Image Text:(e) Mo= 4x X107 T.M/H Calculate the induced emf through the square coil at t = 0.8 s, 1.8 s, 2.8 s, and 3.8 s. (Eand) - Montr²N #1 = (4x × 10^9 T-M/A) (30%/m) x (0.18 m)" (250) do X Calculate the induced current through the square coil, which has a total resistance of 1.28 , as needed to sketch a graph of induced current vs. time, with numerical labels. Take CCW current as positive. 8-1.255= current ↑ flux point of Induced flux Induced current e cu 1.25s 2.955: Current change fen to at constant rate Iind (mA) L 1.0 ( ( flux change from : 0 to Ⓡ induced thy is induced current e cw Slope? GP #4 EM Induction Induced Current vs. Time 2.0 3.0 4.0 → 5.0 time (s) I changes from G flex change from x to. induced Phux is → induced current 5 of 6 си
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